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