An object moves along a straight line and has a constant acceleration of 1.25 cm/s2. At t = 4.38 s, its velocity is -3.8 cm/s. Express your answers in cm/s.

What was the object’s velocity at t = 1.37 s?

To find the object's velocity at t = 1.37 s, we can use the kinematic equation:

v = u + at

Where:
- v is the final velocity
- u is the initial velocity
- a is the constant acceleration
- t is the time

Given:
- Constant acceleration, a = 1.25 cm/s^2
- Velocity at t = 4.38 s, v = -3.8 cm/s
- Time, t = 1.37 s

We need to find the initial velocity (u).

Rearranging the equation, we have:

u = v - at

Substituting the given values into the equation:

u = -3.8 cm/s - (1.25 cm/s^2)(1.37 s)
u = -3.8 cm/s - 1.7125 cm/s
u ≈ -5.5125 cm/s

Therefore, the object's velocity at t = 1.37 s is approximately -5.5125 cm/s.

v(t) = 1.25 + at

so, to find a,
1.25 + 4.38a = -3.8
a = -1.153 cm/s^2
so, v(1.37) = 1.25 - 1.153*1.37 = -0.33 cm/s