Question

A particle starts from rest accelerate at the rate of 4m/s2 for 4sec and then decelerate at the rate of 2m/s2 to come to rest. Find total time taken for the journey and also find displacement of the particle.

Answer 1

Final speed reached at the end of acceleration is given by:
\[\large v=v _{0}+at=4\times4=16\ ms ^{-1}\]
Time taken to decelerate to rest is given by:
\[\large t=\frac{v}{a}=\frac{16}{2}=8\ s\]
Total time =4 + t = 4 + 8 = 12 s
The average speed during acceleration is:
\[\large \frac{0+v}{2}=\frac{16}{2}=8\ ms ^{-1}\]
The average speed during deceleration is:
\[\large \frac{16+0}{2}=8\ ms ^{-1}\]
Therefore the average speed for the whole journey is 8 m/s, and the total time for the journey is 12 s. So the displacement is given by:
\[\large 8\times12=96\ m\]