Motion Worksheet-8
A. 400 metres B. 300 metres
C. 200 metres D. 100 metres
A. 50 metres B. 200 metres
C. 150 metres D. 100 metres
A. 1.2 m/s^{2} B. 1.85 m/s^{2}
C. 2.5 m/s^{2} D. 1.50 m/s^{2}
A. 30 m/s^{2} B. 3 m/s^{2} C. 18 m/s^{2} D. 0.83 m/s^{2}
A. 6.5 m/s B. 5.5 m/s C. 7.5 m/s D. 5.6 m/s
Answer:
Explanation: Here, initial velocity, u = 0
Time, t = 10 s
Acceleration, a = 4 m/s^{2}
And, Distance, s = ?
Now, putting these values in the second equation of motion :
We get :
s = 200 m
thus, the distance covered by the car in 10 seconds is 200 metres.
Explanation: Here, initial speed, u = 10 m/s
Final speed, v = 0 (scooter stops)
Acceleration, a = –0.5 m/s^{2}
Distance covered, s = ?
Now, putting these values in the third equation of motion :
v^{2} = u^{2} + 2a
We get : (0)^{2} = (10)^{2} + 2 × (–0.5) × s
0 = 100 – s
S = 100 m
Thus, the distance covered is 100 metres.
Explanation: Here, initial speed, u = 20 km/h
= 5.55 m/s …(1)
Final speed, v = 60 km/h
= 16.66 m/s …(2)
Acceleration, a = ?
And, Time, t = 6 s
We know that : v = u + at
So, 16.66 = 5.55 + a × 6
6a = 16.66 – 5.55
6a = 11.11
Thus, acceleration, a = 1.85 m/s^{2}
Explanation: Here, initial speed, u = 20 km/h
= 5.5 m/s ………(1)
Final speed, v = 50 km/h
= 13.8 m/s …………(2)
Acceleration, a = ?
And, time, t = 10 s
Putting the value in the equation of motion ;
Now, v = u + at
So, 13.8 = 5.5 + a × 10
10a = 13.8 – 5.5
10a = 8.3
a = (8.3/10)
a = 0.83 m/s^{2}
thus, the acceleration is 0.83 m/s^{2}.
Explanation: We know that for a body moving in a circular path :
v = 2πr/t
Here, speed, v = ?
π = 22/7
Radius of circular track, r = 105 m
And, Time taken for 1 round, t = 2 minutes
= 2 × 60 seconds = 120 s
Now, putting these values of π, r and t in the above formula, we get :
= 5.5 m/s
Thus, the speed of cyclist on the circular track is 5.5 metres per second.