Question

A car accelerates from rest to 60 mph in 4.8 seconds. It then continues at constant velocity for 10 seconds. Determine the distance in the first part and in the second part. Also, what is the car's acceleration in the first part? In the 2nd part? What is it's velocity at the end of the first part( which is also maximum velocity)? (show all work)

distance1=

distance 2=

accel1=

accel2=

velocity max=

Answer 1

Initial velocity of the car = V₁ = 0 m/s (At rest)

Velocity of the car after 4.8 sec = V₂ = 60 mph = 60 x (1609.34/3600) m/s = 26.822 m/s

Time period of the first part = T₁ = 4.8 sec

Acceleration of the car in the first part = a₁

V₂ = V₁ + a₁T₁

26.822 = 0 + a₁(4.8)

a₁ = 5.588 m/s²

Distance covered by the car in the first part = D₁

D₁ = V₁T₁ + a₁T₁²/2

D₁ = (0)(4.8) + (5.588)(4.8)²/2

D₁ = 64.374 m

Time period of the second part = T₂ = 10 sec

The car continues to travel at constant velocity in the second part therefore there is no acceleration.

Acceleration of the car in the second part = a₂ = 0 m/s²

Distance covered by the car in the second part = D₂

D₂ = V₂T

D₂ = (26.822)(10)

D₂ = 268.22 m

a) Distance traveled by the car in the first part = 64.374 m

b) Distance traveled by the car in the second part = 268.22 m

c) Acceleration of the car in the first part = 5.588 m/s²

d) Acceleration of the car in the second part = 0 m/s²

e) Maximum velocity of the car = 26.822 m/s