Question

An object moves at 60 m/s in the +x direction. As it passes through the origin it gets a 4.5 m/s^2 acceleration in the -x direction. a) How much time elapses before it returns back to the origin?

b) What is its velocity when it returns to the origin?

Answer 1

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Step-by-step explanation:

a) Let the starting position be origin and time be t.

After time t displacement, s = 0 m

Initial velocity, u = 60 m/s

Acceleration, a = -4.5 m/s²

We have equation of motion s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

0 = 60 x t + 0.5 x (-4.5) x t²

2.25t² - 60 t = 0

t² - 26.67 t = 0

t (t-26.67) = 0

t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

Initial velocity, u = 60 m/s

Acceleration, a = -4.5 m/s²

Time , t = 26.67

Substituting

v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction