Question

A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 32.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the ride-sharing car in motion (in s)? 62 (b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.) To determine the average velocity, we need to know the total distance traveled and the time to travel this distance. 1648 The total distance traveled consists of three segments-while the vehicle is speeding up, while it is traveling at a constant speed, and while it is slowing down. Did you use the distance and time for all three segments? m/s Need Help? Linen.

Answer 1

Given,

The acceleration, a = 2 m/s²

The speed, V = 32 m/s

We know,

V_f = V_i + at

32 = 0 + 2 * t₁

t₁ = 16 sec

(a)

The total time the ride-sharing car in motion = 16 + 41 + 5

= 62 sec

(b)

The distance, d₁ = u + 1/2 at₁²

   = 0 + 0.5 * 2 * 16²

= 256 m

The distance, d₂ = V * t

= 32 * 41

= 1312 m

For last segment, a = - 32 / 5

= - 6.4 m / s²

d₃ = (0² - 32²) / (2 *(- 6.4))

= 1024 / (12.8)

= 80 m

d = d₁ +d₂ +d₃

= 256 + 1312 + 80

= 1648 m

V = d / t

= 1648 / 62

= 26.58 m/s