What is its displacement (in meters) from the initial moment to t = 7.50s?

Question

An object's initial velocity is 1.34m/s in the +x direction. It slows down with a constant acceleration whose magnitude is 1.49m/s^2. After it reaches a momentary stop it reverses its direction of motion, to the -x, and speeds up with the same magnitude of the acceleration.

Answer 1

As we have uniform accelerated motion, to describe it you need to use the following equation: x(t)=x(0) + v(o)t -+ at²/2  for positive (+x) direction.

 However, according to given data we have deceleration, that is, the acceleration is negative, at first, we will need to find the time at which the object reversed it motion, that is, time at which the velocity is zero.

As v(t)= v(0) – at=0 →t _reverse=1.34/1.49=0.899 s≈0.9 s.

It initial displacement in +x direction was ∆x=x(t) –x(0) = v(o)t -at² /2 =1.34 m/s)(0.9 s) =0.9 m

After that the object continues its motion with the same acceleration and zero initial velocity in –x direction during t=(7.5 s- 0.9 s)= 6.6 s.

∆x(t)=0.9 m -(1.49/2)(6.6) ²= -31.55, that is, displacement 31.55 m to the left

Answer 2

X(t) = V_ot + 1/2at²

X(7.5) = (1.34 x 7.5) + (0.5 x -1.49 x 7.5²)

X(7.5) = -31.85625 ~ -31.9m