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