Answer:
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
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
