since a = 2.7 m/s^2, at t=4.2, v=11.34m/s
Then its acceleration drops to (13.0-11.34)/5 = 0.332 m/s^2
So, starting at t=4.2,
x = -108 + 0.332t + 0.166 t^2
At 5.7s after the first phase, then x = -100.7
A particle travels along the x axis with a constant acceleration of +2.7 m/s2. At a time 4.2s following its start, it is at x = -108 m. At a time 5.7 s later it has a velocity of +13.0 m/s. Find its position at this later time.
Please use kinematic equations to solve.
1 answer
1 answer