Question

Suppose the equation of the velocity of a car at any given time is Vx= 30m/s+(2.5m/s^3)t^2 Find the change in the velocity of the car during time t1= 4s and t2= 5.5s (a) 70.00 m/s (b) 105.63 m/s (c) 35.63 m/s (d) 175.63 m/s

Answer 1

Answer:

time=4, answer choice A

time=5.5, answer choice B

Explanation:

This is taken from the kinematic equation, v=vo+(1/2)at^2

Velocity at time 4 = 30+(2.5)(4)^2

Velocity at time 5.5 = 30+(2.5)(5.5)^2

Velocity at time 4 seconds = 70 m/s

Velocity at time 5.5 seconds = 105.63 m/s