What is the distance travelled by a ball that starts at 2 m/s and moves with an acceleration 4 m/s^2 to reach 18 m/s.
Acceleration is the rate of change of speed. For a body travelling at a speed u, the speed after time t is given by u + at where a is the rate of acceleration. Now the distance travelled by the body is the average of the initial speed and the final speed multiplied by the time travelled.
Here final speed is 18. Now as acceleration is 4,
18 = 2 + 4t
= > t = 4.
Therefore the time taken is 4 sec.
So the distance travelled is given as 4*(2+ 18)/2 = 4* 10 = 40 meters.
The required distance travelled is 40 m.
The initial velocity of the ball, u = 2m/s
The final vecity of the ball , v = 18m/s.
the acceleration a while attaining the final velocit is : a = 4m/s^2.
The distance s moved by the ball is as a function of u , v and a is given by:
2as = v^2-u^2.
2*4*s = 18^2-2^2
8s = 324-4 = 320
s = 320/8 = 40.
Therefore the ball under the given dynamic condtions has covered a sidtance of 40 meter.