Advertisements
Advertisements
Question
A body starts to slide over a horizontal surface with an initial velocity of 0.5 ms−1. Due to friction, its velocity decreases at the rate of 0.05 ms−2. How much time will it take for the body to stop?
Advertisements
Solution
Initial velocity of body u = 0.5 ms−1.
Final velocity of the body v = 0 ms−1 as body comes to rest finally.
Retardation of body = 0.05 ms−2.
We know that v = u + at.
0 = 0.5 − 0.05t
T = 0.5/0.05
T = 10 sec.
RELATED QUESTIONS
A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Multiple choice Question. Select the correct option.
A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km/h to 18 km/h in 5 s. What is the distance travelled by train during this interval of time?
A boy drops a stone from a cliff, reaches the ground in 8 seconds. Calculate
- final velocity of stone
- height of cliff. (Take g = 9.8 ms−2)
A stone thrown vertically upwards takes 4 s to return to the thrower. Calculate
- initial velocity of the stone
- maximum height attained by stone. (Take g = 10 ms−2)
Derive the equations
(i) v= u+at and
(ii) v2-u2= 2as, where the symbols have their usual meanings.
Derive the equation
V2-u2 = 2as
A car moving with a uniform acceleration of 10 m/s2 on a straight road changes its speed from 10 m/s to 30 m/s. Find the time lapsed for the change of speed.
A train is moving at a speed of 90 km/h. On applying brakes, a retardation of 2.5 ms-2 is created. At what distance before, should the driver apply the brakes to stop the train at the station?
