Engineering Principles
Solution 1a
As given in question,
Y = -h, a = -9.8 m/s, t =?, S = -2.0 m, u = 4.5 m/sec
We know that distance travelled is given by
By solving these equations, we get,
T = 1.245 and -0.3276,
Time is non-negative, therefore, her feet in the air = 1.245 sec
Solution 1b
The highest point is calculated through
Putting the requied value from question (a)
Putting the value in this equation,
1.03316 m Ans
Solution 1c
The velocity at water level =
= -0.70701 m/sec
Solution 2a
As given in question, V = 80 m/sec, g = 10 m/sec2, u = 0
We know that
Then, Putting the value
Ans
Solution 2b
Suppose the ball started from ground h = 0, Then time taken to reach the ball at the height of 320 m
Then t = 8 sec
Time taken to reach to the ground = 8*2 = 16 sec Ans
Solution 2c
, u = 80 m/sec, g = 10 m/sec2, u = 0, t = 4 sec
The
Putting the value =
Solution 2d
The maximum acceleration of the ball at maximum height,
The acceleration will be 10 m/s2 downward
Solution 2e
Since there is no external force acting on ball after release, therefore, acceleration = 10m/sec2 upward or downward. The graph is as follows
Solution 2f
Since initial velocity is constant, the final velocity is changing with time, The graph of final velocity vs time is given below
Solution 3a
Suppose the value of acceleration due to gravity is taken as 9.8 m/sec2
H = 30 m, T = 1 second,
We know that, Putting the value in it.
The distance travelled in 1 second = 4.9 m
Solution 3b.
u= 0, g = 9.8 m/s2, h = 30m,
As per equation
then, v = 24.25 m/sec
Solution 3c.
First, we must calculate time taken to reach the ground
Putting the values
t = 2.47
Distance covered in 1.47 sec = 0.5*9.8*1.47 = 10.58 m
Then, distance covered = 30.0-10.58 = 19.41 m
Solution 4a.
The relative velocity of 2nd runner = 4.5-3.75 = 0.75 m/sec
Solution 4b.
Time taken by first runner = 250/3.75 = 66.66 sec
Time taken by second runner = (250+50)/4.5 = 66.66 sec
The time taken to cover their respective distance is equal to 66.66 sec, Therefore, both runners reach at the same time
Solution 5a.
Since both the runner, move reach the finish time at same time, so no gap between them at finish line.
As given in the figure,
Suppose the velocity of jet w.r.t air is given as Vja
And velocity of air w.r.t. ground = Vag
Now velocity of plane w.r.t ground
Then Vjg = ….(i)
Vjg = -219.081i-31.333j
Solution 5b.
The magnitude of this vector is resultant speed = 221.31 m/sec
The direction = Ans
Solution 6a.
Solution 6b.
As given in figure, the equilibrium of motion can be given as for horizontal direction
The constant velocity means, force will be applied only for friction , and acceleration is zero
Or F = 50/0.866 = 57.73 N
Worksheet One – Measurement and Mensuration
Solution 1a
100 km/hour = 1000*100/3600 = 27.77 m/sec
Solution 1b
100 km/hour = 100/1.609 = 62.1371 miles/hour
Solution 2a
33 m/s = 33*3.6 = 118.8 km/hour
Solution 2b
The result is 118.8 km/h, therefore, it is exceeding 90km/h
Solution 3
Since we know that,
Since, distance covered in 1 sec = 1 m
Then, distance covered in 3600 sec = 3600 m
Distance covered in 1 hr = 3600m = 3.6 km
Therefore, 1.0 m/s = 3.6 km/h
Solution 4
1 yard = 3ft, and 3.281 ft = 1 m
Then 100 yard = 300ft = 300/3.281 = 91.435 m Ans
Solution 5
Length of the Soccer field = 115 m = 115*3.281 = 377.315ft = 377 ft and 0.315*12 = 3.78 inch
Length of the Soccer field = 377’ 3.78” Ans
Width of the Soccer field = 85 m = 85 x 3.281 = 278.885 ft = 278 ft and 0.885*12 = 10.61 inch
Width of the Soccer field = 278’10.61”
Solution 6
6 ft 1inch = 72+1 = 73 inch,
Solution 7
As given in question,
3281 ft = 1 km
Then, 29028 ft = 29028/3281 = 8.8473 km
Solution 8
As given in question
Distance in 1 sec = 342 m/s
3600 sec = 342*3600 = 1231200 m = 1231200/1000 = 1231.2 km /hr Ans
Solution 9a
As given in question
1 year = 24*364*60*60 seconds move to 4 cm
1 sec = 4/3153600 = 126839 x 10-7 cm /sec Ans
Solution 9b
As given in question,
In 1 year moves 0.00004 km
Then 1000000 yrs. moves = 0.00004*100000 = 40 km/million-year Ans
Solution 10
As given in figure, distance between earth and sun = r = 108 = 10000000 km
Assuming as circular orbit, the distance covered by earth in 1 revolution
Putting the values,
Distance travelled by earth in 1 year (31536000 sec) =
Distance travelled by earth in 1 second = Ans
Solution 11
The uncertainty for a quantity is given as
The uncertainty of mas in 65 kg is 1.95 kg
Solution 12
The % uncertainty in measuring (0.5 cm = 0.005 m)
Ans
Solution 13
The uncertainty for a quantity is given as
The uncertainty of mas in 90 km is 4.5 km, which means the actual speed is in the range of 90-4.5 = 85.5 km/h or 90+4.5 = 94.5 km/h
Or, range is 85.5*0.6214 = 53.13 miles/hour to 58.72 miles/hour
Solution 14
The % uncertainty in measuring
Ans
Solution 15
As given in question, 72 beats /min = 72/60 = 1.2 beats /sec (Since 1 year = 31536000 sec)
In 2.0 years, the no of beats = 31536000*1.2*2 =75686400 = 7.5 x 107 beats
In 2.00 years, the no of beats = 75.68x 106 beats
In 2.000 years, the no of beats = 7.568 x 107 beats
Solution 16
The Soda left after 308 ml removal = 375-308 = 67 mL
Solution 17a
In the given equation the lowest number of significant figures is 3.
Therefore, answer should be 3 significant figures
Solution 17b
(18.7)2 Only one number which has significant figure 3. Therefore, solution will be 3 significant figures.
Solution 17c
(1.60 x 10-19) (3712) = The lowest significant figure is 3, Therefore solution should be 3 significant figures.
Solution 18a
The number 99 consists of 2 significant figures, and 100 has 3 significant figures
Solution 18b
The uncertainty in 99 =
The uncertainty in 100 =
Solution 18c
As per result obtained from part a and b, The % uncertainty show deviation from the results, therefore, % uncertainty is more meaningful.
Solution 19a
The % uncertainty in speedometer = = 2.22%
Solution 19b
The range of speed at 60 km/h =
The range of the speed in between = 55.667 to 61.333 km/h Ans
Solution 20a
The % uncertainty in blood pressure= = 1.667%
Solution 20b
The uncertainty in blood pressure at the 80 mmHg = 80*1.667/100 = 1.33 mm Hg
Solution 21
The heart rate for one minute =
The % uncertainty in heart rate= = 2.5% + 1.7% = 4.2%
Therefore, uncertainty in heart rate
Therefore, heart rate is 80 beats/min with uncertainty 3 beats/min
Solution 22
The area of circle is given by = Then putting the values
Ans
Solution 23
The time taken to cover the marathon at the speed of 9.5 mi/h Ans
Solution 24
The distance in meter = 42188 m
Solution 24a
The uncertainty in distance =
Solution 24b
Time in seconds = 2*3600+30*60+12= 9012 sec Ans
The uncertainty in time = Ans
Solution 24c
The average speed in meter / sec = 42188/9012= 4.681 m/sec Ans
Solution 24d
In this condition we must take two averages as per uncertainty, one is max and other is min average.
Max average
Min Average = m/sec Ans
Solution 25
The volume of the box can be given as (Putting the value)
Volume of box = 1.8 x 2.05×3.1 = 11.439 cm3
The max volume with uncertainty = (1.8+0.01) (2.05+0.02) (3.1+0.1)
Vmax = 11.99 cm3
The min volume with uncertainty = (1.8-0.01) (2.05-0.02) (3.1-0.1)= Vmin = 10.90 cm3
The volumetric with uncertainty = 11.99-10.90 = 1.09 cm3 Ans
Solution 26
Since 1 pound = 0.4539 kg
Then, 0.0001kg = 0.0001/0.4539 = 0.00022 lb
The % uncertainty in pound = 0.00022/1 % = 0.022 %
The uncertainty in 1kg = 1000 gm = 1000*0.022 = 22 gm Ans
Solution 27
The area of room A = lxb = 3.955*3.050 = 12.063 cm2
The Max area of room with uncertainty = (3.955+0.005) (3.05+0.005) = 12.098
The min area of room with uncertainty = (3.955-0.005) (3.05-0.005) = 12.028
The amount of uncertainty in area = 12.098-12.028 = 0.070 cm2 Ans
Solution 28a
The volume of piston =
The gas decreases in piston in volume = 143.508 cm3
Solution 28b
The max volume with uncertainty = = 143.63 cm3
The max volume with uncertainty = = 143.39 cm3
Uncertainty in volume = 143.63-143.39 = 0.2414 cm3
Worksheet Two – Kinematics
Solution 1a
The distance covered by path A = Ans
Solution 1b
The magnitude of the distance covered = Ans
Solution 1c
The displacement from start to finish = Ans
Solution 2a
The distance covered by path B = Ans
Solution 2b
The magnitude of the distance covered = Ans
Solution 2c
The displacement from start to finish = Ans
Solution 3a
The distance covered by path C
=
Ans
Solution 3b
The magnitude is given by = Ans
Solution 3c
The displacement from start to finish = Ans
Solution 4a
The distance covered by path D
= Ans
Solution 4b
The magnitude of the distance covered = Ans
Solution 4c
The displacement from start to finish = Ans
Solution 5a
As given in figure, distance between earth and sun = r = 108 = 10000000 km
Assuming as circular orbit, the distance covered by earth in 1 revolution
Putting the values,
Distance travelled by earth in 1 year (31536000 sec) =
Distance travelled by earth in 1 second =
Solution 5b
Since the average velocity in a year = displacement /time,
But in case of revolution displacement is zero, therefore, average velocity = 0
Solution 6
Since the average velocity in a year = displacement /time,
But in case of revolution displacement is zero, therefore, average velocity = 0
Solution 6
As given in question, 100 revolution /min = 100/60 = 5/3 rev /sec
The circumference of the radius = = 2 *3.14*5 = 31.4 m
Distance covered in 1 sec = speed = 31.4*5/3 = 52.33 m/sec Ans
Solution 7
500 km = 500*1000*100 cm
Speed of drifting = 3 cm/year
Time taken to cover 5.0*107 cm = 5/3 * 106 = 1.66 x 107 year Ans
Solution 8
590 km = 590*1000*100 cm
Speed of drifting = 6 cm/year
Time taken to cover 5.90*107 cm = 5/3 * 106 = 9.833 x 106 year Ans
Solution 9
Total time in hour = 13h + 4/60 h+ 58/3600 hour = 13.08278 h
Average speed of train = distance / time = 1633.8/13.08278 = 124.88 km/h
Speed in m/sec = 124880/3600 = 34.68 m/sec
Solution 10
We know that, 3.84 x 106 m = 384×106 cm
Time taken to cover this distance = 96 x 106 year Ans
Solution 11a
The average speed of the car = 12*60/18 = 40 km/hour
Solution 11b
The average velocity = displacement / time = 10.3 *60/18 = 34.33 at the direction 25o south of east.
Solution 11c
Total time by trip = 7.5 hr
Overall speed of return journey = 24/7.5 = 3.2 km /hr.
Solution 11d
Since the overall displacement is zero, therefore, average velocity will also be zero
Solution 12
Time taken to travel the nerve cell = 1.1/18 = 0.0611 sec Ans
Solution 13
Distance covered by echo time = distance between earth and moon = speed x time
Distance between earth and moon = 2.56 x 3 x108 = 7.68 x 108 meter Ans
Solution 14a
The velocity for straight down 15 m = 15/2.5 = 6
The velocity of backward 3 m =3/1.75 = 1.71 m
The velocity of straight forward = 21/5.2 = 4.0384 m/sec
Solution 14b
Total displacement = 15 -3+21 = 33 m
Total time = 2.5 + 1.75 + 5.2 = 9.45 sec
Then average velocity = 33/9.45 = 3.49 m /sec in forward direction
Solution 15a
Circumference of the hydrogen atom = 3.14 x 1.06 x 10-10 m,
Speed of the electron = 2.2 x 10-10 m/s
Time in one revolution = = 1.5129 x 10-16 sec
Revolution in one sec = 1/(1.5129 x10-15) = 6.61 x 1015
Solution 15b
Since displacement is zero after each revolution, then velocity will also be zero.
Solution 16
We know that acceleration = v-u/t = 30-0/7 = 4.286 m/sec2 Ans
Solution 17a
For first case, v = 282 m/s, u = 0, t = 5 s
Acceleration = 282-0/ 5 = 56.4 m/sec2 = 56.4/9.8 = 5.755 g
Solution 17b
For second case, u = 282 m/s, v = 0, t = 1.4
Then deceleration = 0-282/1.4 = -201.42 m/sec2 = -201.42/9.8 = 20.55g Ans
Solution 18a
As given in question, v = 2.00 m/s, u = 0, a = 1.4 m/s2
V =u+at = 0+1.4t, or t = 2.00/1.4 = 1.4285 seconds
Solution 18b
U = 2.00, v = 0, t = 0.8 sec
Then a = v-u/t = 0-2/0.8 = -2.5 m/s2 Ans
Solution 19
As given in question, 6.5 km/s = 6.5*1000 = 6500 m/sec t = 60 s, u = 0
Then acceleration = v-u/t = 6500-0/60 = 108.33 m/s2
The average acceleration of ballistic missile = 108.33/9.8 = 11.054 g Ans
Solution 20
As given in question,
A = 4.50 m/s2, t = 2.4 s, u = 0
Then v = u+at , v = 0+4.5*2.4 = 10.8 m/s2 Ans
The graph of distance vs time is given below
Ans
Solution 21
As given in question,
v = 0, a = -2.1 x 104 m/s2, u = ?, t = 1.85 x 10-3 seconds
We know that, v = u+at, Putting the values,
0 = u – 2.1 x 104 x 1.85 x 10-3 or, u = 2.1 x 104 x 1.85 x 10-3 = 38.85 m/s
Velocity when touches the mitt = 38.85 m/s
Solution 22
As given in question,
U = 0, t = 8.1 x 10-4, a = 6.2 x 105 m/s2, v = muzzle velocity = ?
We know that v = u+at, then v = 0+ 8.1*10-4 x 6.2×105 = 502.2 m/s
Therefore muzzle velocity = 502.2 m/s
Solution 23a
As given in question,
U = 0, t =?, a = 1.35 m/s2, v = 80*1000 m /h = 22.22 m/s
We know that v = u+at, then 22.22 = 0 + 1.35*t
Time taken to accelerate = t = 22.22/1.35 = 16.46 sec
Solution 23b
As given in question,
U = 22.22, t =?, a = -1.65 m/s2, v = 0
V = u+at, t = v-u/a = -22.22/-1.65 = 13.46 seconds
Solution 23b
During emergency
U = 22.22, t = 8.3, v = 0, a =?
We know that, putting the values,
In case of emergency train decelerates at -2.67 m/2
Solution 24a
In this problem,
Solution 24b
The known variable are listed below
U = 12 m/s, a = 2.4 m/s2, t = 12 seconds, and unknown is final velocity v
Solution 24c
The unknown variable is final velocity, which can be calculated as follows
Putting the values
V = 0+2.4 x 12 = 28.8 m/sec.
After knowing the unknown variable we can calculate the distance travelled in 12 seconds, which is as follows
Where S is the distance travelled in meter. Putting the value in above equation,
28.82 = 02+2 x 2.4 x S
Or, S = 28.82/4.8 = 172.8 m. therefore, distance travelled = 172.8 m
Solution 25a
U = 9.00 m/s, a = 2.00 m/s2, t = 5.00 sec
Distance travelled in 5 sec = ut +0.5at2 = 9*5-0.5*2*52 = 45-25 = 20 m
Therefore, distance travelled by runner = 20 m
Solution 25b
Final velocity after 5 sec, v = u+at, v = 9-2*5 = 9-10 = -1 m/sec
The calculation shows that, final velocity is negative, we must calculate the time at final velocity 0
T = v-u/a = 0-9/2 = 4.5 sec,
Solution 25c
The runner stops at 4.5 sec, therefore final velocity of the runner must be zero after 5 sec
Ans
Solution 26a
The sketch is given below
Solution 26b
The known variable in this problem is listed below
U = 0, v = 30 cm/s, S = 1.8 cm
Solution 26c
As given above known variable,
Putting the value in this equation
302 = 0+2 x a x 1.8
Then, a = 900/3.6 = 250 cm/sec2,
Then time taken to accelerated = t = 30/250 = 0.12 seconds
Solution 26d
Since our heart rate is 72 beats in 60 seconds, i.e. one beat is in 0.83 seconds, The time of blood acceleration is only a fraction of time, therefore, answer is reasonable.
Solution 27
U = 8 m/s, v = 40 m/s, t = 3.33 x 10-2 s, a = ?, S = ?
We know that, a = (v-u)/t = (40-8 )/3.33×10-2 = 960.961,
The distance covered during acceleration
Putting the value in this equation
= 402 = 82 + 2 x 960.961xS
Then S = 1536/1921.922 = 0.7992 m Ans.
Solution 28
U = 0, v = 26.8 m/s, t = 3.9 s, a =?
The average acceleration of bike = 26.8-0/3.9 = 6.87 m/s2
Distance travelled during acceleration
S = 0.5 x 6.87 x 3.92 = 52.25 m Ans
Solution 29a
As given in question,
U = 4 m/s, a = 0.05m/sec2, v = ?, t= 8*60 = 480 sec
We know that, v=u+at, Putting the value in eq.
V= 4+480*0.05 = 24+4 = 28 m/sec
Final speed of train is 28 m/s
Solution 29b
U = 28 m/s, v = 0, a = .55m/s, t = ?
We know that, =( 28-0)/0.55 = 50.9090 seconds
Time taken to stop the train 50.9090 seconds
Solution 29c
Case 1 Acceleration,
Now, u = 4 m/s, t = 480 s, a = 0.05
We know that
S = 4 * 480 + 0.5 x 0.05×4802 = 7680 m
Case 2 deceleration,
Now, u = 28 m/s, t = 50.91 s, a = 0.55
We know that
S = 50.91 * 28 – 0.5 x 0.55×50.912 = 712.7273 m
Solution 30a
U = 0, v = 65.0 m/sec, S = 0.250 m, a =?, t = ?
Putting the value in this equation
= 652 = 02 + 2 x 0.25xa
a = 4225/0.5 = 8450 m/s2
U = 0, v = 65.0 m/sec, S = 0.250 m, a = 8450 m/s2, t =?
T = (v-u)/a = 65/8450 = 0.007692 s Ans.
Solution 30b
U = 0, v = 65.0 m/sec, S = 0.250 m, a =?, t = ?
Putting the value in this equation
= 652 = 02 + 2 x 0.25xa
a = 4225/0.5 = 8450 m/s2
Solution 31a
U = 0, v = 6 m/s, a = 0.35 m/s2
Distance travelled in accelerating
62 = 02 + 2×0.35xS
Then S = 36/0.7 = 41.42 m
Solution 31b
V = 6 m/s, u = 0, a = 0.35 m/s2, t =0
T = (v-u)/a = 6/0.35 = 17.14 seconds
Therefore, time taken to become airborne is 17.14 seconds and distance covered is 41.42 m
Solution 32a
As given in question,
U = 0.60 m/s, v = 0, S= 2.00 mm = 2 x 10-3 m
We know that, = -90 m/s2
The deceleration in g format = 90/9.8 = 9.184 g
Solution 32b
Now time can be calculated as
Putting the values
0.00667 seconds Ans
Solution 33a
U = 7.5 m/s, v = 0, S= 0.35 m,
We know that, = -80.36 m/s2 Ans
Solution 33b
Now time can be calculated as
Putting the values
0.0933 seconds Ans
The deceleration of player -80.36 and time to collide = 0.0933 s
Solution 34
As given in question,
U = 54m/s, g = -9.8 m/sec2, v = 0, t=? Sec, S = 3 m
We know that, = -486 m/s2
Solution 35a
As given
V = ?, u = 0, g = 9.8 m/s2, h = 3 m
Putting the value in this equation
V2 = 2 x 9.8 x 3 , then v = 7.668 m/s
Solution 35b
Now, u = 7.668 m/sec, v = 0, S = 2/100 = 0.02 m
We know that, = -1469.96 m/s2
Therefore, squirrel decelerates at 1469.96 m/s2
Solution 36a
As given, u = 22 m/s, a =- 0.15 m/s2, S =210 m,
The time taken to travel the nose of the train will be distance covered by engine on the platform
S = ut +0.5at2 Putting the value in equation
210 = 22t-0.5×0.15t2,
Then 0.075t2 – 22t+210 then t = 9.8781, 283.46
Therefore, t = 9.8781 and 283.46 Ans.
To decide the time of crossing the train, we must find the time to become stops the train
Using V = u + at, Then, 0 = 22-0.15*t, or t = 22/0.15 = 146.667.
Since the train stops at 146.667 seconds, therefore, train must cross the platform at time 9.87 s
Solution 36c
The distance travelled by train must include platform + length of train = 130+210 = 340m
We will use the following station
Putting the values, 340 = 22t- 0.5*0.15*t2
Or, 0.075t2 -22t +340 = 0 , T = 16.37 seconds Ans
Solution 36d
U = 22 m/sec, S = 340 m, a = -0.15 m/s2
Using the formula,
Ans
Solution 37a
The average acceleration of the dragster Ans
Solution 37b
As information obtained from (a)
U = 0 m/se, S = 402 m, v =?, a = 32.58 m/s2
= + 2 *32.58*402
Then v = 161.8466 m/sec Ans
Solution 37c
For any vehicle, the change of gear reduces the acceleration of the vehicle, same has happened to dragster also. The final velocity would be less that what calculated above.
Solution 38a
U = 11.5 m/s, a = 0.5 m/s2, t = 7 sec
Solution 38b
Time taken to cover 300 m at constant velocity 11.5 m/s = 300/11.5 = 26.09 s
Distance covered at 0.5 m/s2 acceleration = S = 11.5 x 7 +0.5×0.5*7 = 92.75
The distance covered by racer in 7 seconds is 92.75
The rest distance covered by constant velocity of 15 m/sec = 300-92.75 = 13.82
Total time taken by racer = 13.82+7 = 20.82
Saved time = 26.09-20.82 = 5.27 Seconds
Solution 38c
Th time taken by 2nd racer = 295/11.8 = 25.0 s
Time taken by first racer to cover 300 m = 20.82 sec
The time difference of winner racer = 25-20.82 = 4.18 seconds Ans
The difference in distance = 4.18*15 = 62.7 m
Solution 39
The acceleration can be calculated as,
The time required to attain the speed = 183.58 mi/h
The distance travelled during acceleration
The distance remained to cover = 5.0-0.31 = 4.7 mi
And Time for rest of the distance =
Then total time required to cover the distance = 92.17 + 12.2 = 104.37 seconds
Solution 40a
A given in question u = 0, t = 9.69 sec, time for acceleration = 3 s, S = 100 m
The distance covered in constant speed =
If we substitute the value of d for velocity part
Putting the value d =1.5v
100-1.5v= 6.69v
Or, v = 100/8.19 = 12.21 m/s
Now we got final velocity v =12.21 m/s
Then acceleration
Ans
Solution 40b
In this case the decelerating part is = 200-d
The covered distance = d =
Equating the problem as previous
200 – d = v(16.30)
Or, v = 200/17.88 = 11.2 m/se
Therefore, maximum velocity of Usain Bolt = 11.2 m/se
Solution 41a
As given in question,
U = 15.0 m/s, g = -9.8 (Upward), t = 0.5
The attained height H = 15*0.5 – 0.5*9.8*.52 = 7.5 – 1.225 = 6.275 m
Final velocity v = 15-9.8*0.5 = 15-4.9 = 10.1 m/s
Solution 41b
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 1.00
The attained height H = 15*1 – 0.5*9.8*12 = 15 – 4.9 = 10.1 m
Final velocity v = 15-9.8*0.5 = 15-4.9 = 5.2 m/s
Solution 41c
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 1.50
The attained height H = 15*1.5 – 0.5*9.8*1.52 = 22.5 – 11.025 = 11.475 m
Final velocity v = 15—14.7= 0.3 m/s
Solution 41d
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 2.0
The attained height H = 15*2 – 0.5*9.8*22 = 30 – 19.6 = 10.4 m (falling)
Final velocity v = 15—19.6= -4.6 m/s (downward)
Solution 41e
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 2.5
The attained height H = 15*2.5 – 0.5*9.8*2.52 = 37.5 – 30.625= 6.875 m (falling)
Final velocity v = 15—24.5= -9.5 m/s (downward)
Solution 42a
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 0.5, H = 70 m
The attained height H = 14*0.5 + 0.5*9.8*.52 = 7 – 1.225 = 8.225 m
Position of stone = 70-8.225 = 61.775 m above ground
Final velocity v = 14+9.8*0.5 = 14+4.9 = 18.9 m/s
Solution 42b
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 1.0, H = 70 m
The attained height H = 14*1 + 0.5*9.8*12 = 14 + 4.9 = 18.9 m
Position of stone = 70-18.9 = 51.1 m above ground
Final velocity v = 14+9.8*0.5 = 14+9.8 = 23.8 m/s
Solution 42c
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 1.5, H = 70 m
The attained height H = 14*1.5 + 0.5*9.8*1.52 = 32.025 + 37.975 = 37.975 m
Position of stone = 70-32.025 = 37.975 m above ground
Final velocity v = 14+9.8*1.5 = 14+14.7 = 28.7 m/s
Solution 42d
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 2.0, H = 70 m
The attained height H = 14*2.0 + 0.5*9.8*2.02 = 28 + 19.6 = 47.6 m
Position of stone = 70-47.6 =22.4 m above ground
Final velocity v = 14+9.8*2.0 = 14 19.6 = 33.6 m/s
Solution 42e
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 2.5, H = 70 m
The attained height H = 14*2.5 + 0.5*9.8*2.52 = 35 + 30.625 = 66.625 m
Position of stone = 70-4.375 =4.375 m above ground
Final velocity v = 14+9.8*2.5 = 14 +24.5 = 38.5 m/s
Solution 43
U = 0, g = -9.8 m/s2, H = 1.25 m
We know that, Putting the value in equation
Ans
Solution 44a
The known variable is as follows
U = 1.4 m/s, t = 1.8 s, g = 9.8 m/s2
Solution 44b
H = ut+0.5t2 = 1.4×1.8 + 0.5*1.8*1.8*9.8 = 2.52 + 15.876 = 18.396 m Ans
Solution 45a
The known variable in this problem is as follows
U = 13.0 m/s, g = 9.8 m.s2, v = 0, H = ?
The height attained by dolphin, is at hat point where final velocity gong upward will be zero.
Solution 45b
From the equation
Putting the value of known variable
0 = 132 -2 *9.8*H, or H = 169/19.6 = 8.62245 m
Solution 45c
The time taken to reach the highest point t =
The time taken to going upward is 1.327 s, similarly time taken to going downward = 1.327
Total time in air = 1.327+1.327 = 2.653 s
Solution 46a
As given in question, u = 4 m/s, g = -9.8 m/s2(upward), H = 1.8 m, v = 0
Since The swimmer jumps from board, therefore, distance covered will addition of board height
0 = 42 – 2*9.8 *H
Or H = 16/19.6 = 0.816 m,
The total height = 1.8 m+0.816 = 2.616 m
The time taken in going up = t = 4-0/9.8 = 0.408 s
Now time taken to come to the water
S = ut+1/2gt2 Putting the value,
2.616 = 4.9t2 or, = 1.37 sec
Total time in the air = 0.408+1.37 = 1.77661 s
Solution 46b
As given in question, u = 4 m/s, g = -9.8 m/s2(upward), H = 1.8 m, v = 0
Since The swimmer jumps from board, therefore, distance covered will addition of board height
0 = 42 – 2*9.8 *H
Or H = 16/19.6 = 0.816 m,
The total height = 1.8 m+0.816 = 2.616 m
Solution 46c
The final velocity can be calculated as
V2 = 2 * 9.8*2.616 = 51.2736, then v = 7.1606 m/sec Ans
Solution 47a
As given, t = 2.35 s, u = 8.00 m/s, g =-9.8 m/s2 (Upward)
H = ut+0.5gt2 = 8*2-0.5*9.8*2.352 =
H = ut+0.5gt2 = 8*2+0.5*9.8*2.352 = 18.8 + 27.0602 = 8.26 m
The height of the Thrown stone from cliff = 8.26m
Time taken to reach the top point
T = -8/-9.8 = 0.816 s
The time taken by stone to reach the ground = 2.35 – 0.816 = 1.533673
Now I must calculate distance covered in 1.534 sec
H = ut +0.5gt2 = 0+0.5*9.8*1.5342 = 11.52556
Then height of the cliff = 3.265 m
Solution 47b
Time taken to reach the stone when thrown downward,
U = 8.0 m/s, H = 3.265, g = 9.8
We know that, H = ut +0.5at2, Putting the value
3.265 m = 8t+4.9t2
4.9t2 + 8t-3.265 = 0
Calculated root is
T = 0.33811 and – 1.97076
Since negative amount is neglected for time, t = 0.33811 seconds Ans
Solution 48
As given in question,
U = 11 m/sec, g = 9.8 m/sec2, height = 2.2m, another height = 1.8 m
The equation for distance covered = H = H’ + ut +0.5at2
Putting the value
2.2 = 1.8 +11 -0.5 *9.8*t2
Solving the equation through quadratic equation
Negative value is omitted, t = 2.26 seconds,
Then shot putter must get away = 2.26 seconds
Solution 49
As given, u = 15 m./sec, H = 7m, g = 9.8,
The formula for distance
With the help of above equation, the equation of time
Putting the required value from question,
After solving the above quadratic equation, we get t =0.57 and 2.49 s
Tome taken to pass the branch while upward = 0.57 s
Tome taken to pass the branch while upward = 2.49 s
The required time interval = 2.49 -0.57 = 1.92 s Ans
Solution 50a
As given, H = 2.5 m, v =0, u =?, g = 9.8 (upward)
Putting the value
02 = u2 – 2*9.8*2.5, u2 = 49, then u = 7 m/s Ans
Solution 50b
Time taken to reach the air = t = 2H/(u+v) = 5/7 = 0.714 seconds Ans
Solution 51a
The reference plane distance = 0, Suppose the rock reaches at place H after 1.5 s
Then
H-Ho = ut+1/2gt2, y = 0*1.5 + ½(-9.8)(1.5)2 = -11.02 m
The rock will be visible for hiker H = 105 – 11.02 = 93.98 m
Solution 51b
The time taken to reach the ground by rock can e given as
Putting the values
Time for hiker to take action = 4.6-1.5 = 3.1 seconds Ans
Solution 52a
As given in question
U = 0, t = 1 sec, g = 9.8 m/sec2 (downward)
We know that,
, H = ut +0.5at2, Putting the value
H = 0-4.912 = -4.9 m
Distance traveled in 1st sec = 4.9 m
Solution 52b
The final velocity of the object is
The velocity at which the ball reach at ground = 38 m/sec
Solution 52c
We know that,
or Ans
Then distance travelled by body in first 2.9 second
H = 0 + 0.5*9.8*2.92 = 41 m
Hence distance travelled in last second = 75-41 = 34 m Ans
Solution 53a
As given, u = 0, H = 250 m, g = 9.8 m/sec2(Upward)
The final velocity while hitting the ground = 70 m
Solution 53b
Sound speed = 335 m/s, H = 250
Time taken to reach to sound =250/335 = 0.75 seconds
The reaction time by tourist = T = 0.75 + 0.30 = 1.05 seconds
Time taken by rock to reach = t = (v-u)/a = 70/9.8 = 7.143 seconds
Then time for action taken by tourist = 7.143 – 1.05 = 6.10 seconds Ans
Worksheet Three – Kinematics
Solution 1a
As given in figure,
The distance covered by A is three blocks to the north and one block east,
Then, distance travelled by A = (120 +120+120) +(120) = 480 m Ans
Solution 1b
The displacement is measure as perpendicular = 120 + 120 +120 = 360 m
Now we must calculate hypotenuse
The direction = Ans
Solution 2a
As given in figure,
Distance travelled in east direction, d1 = 4 *120 = 480 m
Distance travelled in north direction d2 = 3 *120 = 360 m
Distance travelled in west direction d3 = 3 *120 = 360 m
Total distance travelled = 480 + 360 + 360 = 1200m = 1.2 km
Solution 2b
Net movement in east = 120 m
Net movement in north = 360 m
The displacement can be given by h =
The direction of the movement = Ans
Solution 3
The displacement component in north direction can be given as
Putting the value
Sn = 5 x Sin40o = 5 *0.642 = 3.21 km
Solution 4
Suppose,
Resultant vector at 90o
The angle between these two vectors
Actual angle = 90o 54.250 = 35.75o
Therefore, the magnitude and direction = 30.8 m and 35.75o west of north
Solution 5
The figure for the arrangement is as follows.
Suppose x component vector = As =A*Cosθ
Putting the value for given θ
As = 12.0 x Cos (90o +20o)
As = 120 x Sin(20o) = 12.0 * (-0.342) = -4.104 m
For the same way, component along y axis
Ay = 12.0 x Cos20o = 12.0 x 0.939 = 11.27 m
Similarly, for B vector component along the x and y axis
Bx = 20 *Cos(220o) = 20 * (-0.667) = -15.32 m
And By = 20 *Sin(220o) = 20 * (-0.6427) = -12.85 m
Therefore, resultant force in x and y axis
Rx = -4.104 – 15.32 = -19.42 m
Ry = 11.27-12.85 = -1.58 m
The resultant magnitude
R = Ans
And direction Ans
Solution 6
The diagram for the problem can be stated as
We must calculate the x component
The resultant R can be calculated as
Ans
Solution 7a
The component in Ax and Ay can be given as
Suppose x component vector = As =A*Cosθ
Putting the value for given θ
As = 12.0 x Cos (90o +20o)
As = 120 x Sin(20o) = 12.0 * (-0.342) = -4.104 m
For the same way, component along y axis
Ay = 12.0 x Cos20o = 12.0 x 0.939 = 11.27 m
Similarly, for B vector component along the x and y axis
Bx = 20 *Cos(40o) = 20 * (-0.667) = 15.32 m
And By = 20 *Sin(40o) = 20 * (-0.6427) = 12.85 m
Therefore, resultant force in x and y axis
Rx = -4.104 +15.32 = 11.21 m
Ry = 11.27+12.85 = 24.12 m
The resultant vector
The resultant magnitude
R = Ans
And direction Ans
Solution 7b
The component in Ax and Ay can be given as
Suppose x component vector = As =A*Cosθ
Putting the value for given θ
As = 12.0 x Cos (90o +20o)
As = 120 x Sin(20o) = 12.0 * (-0.342) = -4.104 m
For the same way, component along y axis
Ay = 12.0 x Cos20o = 12.0 x 0.939 = 11.27 m
Similarly, for B vector component along the x and y axis
Bx = 20 *Cos(220o) = 20 * (-0.667) = -15.32 m
And By = 20 *Sin(220o) = 20 * (-0.6427) = -12.85 m
Therefore, resultant force in x and y axis
Rx = -15.32 –(-4.104) = -11.21 m
Ry = -11.27-12.85 = -24.12 m
The resultant vector
The resultant magnitude
R = Ans
And direction Ans
Solution 8
Suppose there are three two dimensional vectors,
Now Adding three vectors, like X+Y+Z
X+Y+Z = ( ( +(
And adding these vectors in different order
Z+Y+X = ( + (
= ( ( +(
In this condition we can say that, sum of these vector in different order results the same
Solution 9
The rectangular form of vector
Similarly
Adding both vector
R =
The resultant magnitude
The direction can be given as
Angle w.r.t x direction
= 180-89.9o = 90.1o Ans
Solution 10
If we resolve the vector into x and y component, then vector in x direction can be given as
………(i)
Similarly, in y direction
………. (ii)
If we eliminate VB from equation (i) and (ii)
Similarly, for equation (ii),
Equating the above two equation
Therefore, Now putting the values
Ans
Now putting the value of VA and VTotal
Ans
Solution 11
The component of VTotal can be given as
Ans
Solution 12
The resultant component of rotated axis can be given, the x component is given as
Ans
And the y component
Solution 13a
As given in figure, 3.58
The distance traveled by C is given as
X = 1*120+5*120+2*120+1*120+1*120+3*120 = 1560 m = 1.56 km
Solution 13b
Resolving the component in x and y direction,
Then resultant will
The angle can be given as
Ans
Solution 14
The distance travelled along the path D
S = 2*120+6+120+4*120+1*120 = 1560 m
If we resolve the component into x and y direction
And the direction
Solution 15
Resolving the vectors in x and y components
Ans
Solution 16
Resultant
Putting the values
R =
The direction The compass reading will be 90 – 54.25 = 35.75o west to north. Analytical technique is more accurate process.
Solution 17
As given in question,
R = X+Y = Y+X
The resultant
And direction
In this condition we must take alternate path where there is not obstacle.
Solution 18a
As given in fig.
We can see that, <ABC = <BAD
Segment AC is in the direction east
The distance AC
Putting the value in it
AC = 7.5 Sin15o = 1.94 km
Similarly CB = 7.5*Cos15o = 7.24 km
Solution 18a
In this condition the overall route must be ADB. In this condition we can say that, the resulting displacement is still same as AB.
Solution 19a
As per figure the resultant displacement
R = X+Y = X-Y = 18i-25j
And its magnitude
And direction
Solution 19b
As per figure the resultant displacement
R = X+Y = X-Y = -18i+25j
And its magnitude
And direction Ans
Solution 20
As per the vector diagram,
The sum of vector = A+B+C = 0
Or C =-A-B
Resolving the vector into x and y component
The resultant vector C
And its magnitude
And direction Ans
Solution 21a
The component along south direction
Ds = RSinθ = 32Sin(35o) = 18.4 km
DW = RCosθ = 32Cos(35o) = 26.2 km
Solution 21b
Similarly, distance towards South and north direction is SSW and SNW
SSW = RCosθ = 32Cos(10o) = 31.5 km
SNW = RSinθ = 32Cos(10o) = 5.56 km Ans
Solution 22a
As given in diagram, we must resolve all the vector into x and y component
As per the vector diagram,
The sum of vector = A+B+C+D = 0
Or D =-A-B-C
Resolving the vector into x and y component
The resultant vector C
And its magnitude
And direction Ans
Solution 23
The given vector can be Summerside as follows
- i) 5 km at 135o
- ii) 7 km at 300o
iii) 1.3 km at 205o
- iv) 1 km at 0o
- v) 7 km at 85o
- vi) 2 km at 235o
vii) 2.8 km at 10o
The given problem can be represented as in fig
After resolving the given vector in question into x and direction,
The resultant can be given as
And direction Ans
Worksheet Four – Kinematics
Solution 52
