Question

The acceleration of a bus is given by a_x(t)=At, where A= 1.21 is a constant.

A)if the bus's velocity at time t_1= 1.14s is 5.03m/s, what is its velocity at time t_2= 2.04 ?

B)if the bus's velocity at time t_1= 1.14s is 6.02m/s, what is its velocity at time t_2= 2.04 ?

Answer 1

A) The velocity change is the integral of the acceleration vs time

change in V = INTEGRAL 1.21 t dt
from t = 1.14 to t = 2.04 s
= 1.21[(2.04)^2 - (1.14)^2]/2

B) Same approach; same velocity change; different starting velocity

Answer 2

1.73