Question

A taxi traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 27.0 m/s. Then the vehicle travels 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 taxi in motion (in s)? s (b) What is the average velocity of the taxi for the motion described? (Enter the magnitude in m/s.) m/s

Answer 1

Part a)

`T = 59.5 s`

Part b)

v = 22.8 m/s

Step-by-step explanation:

Part a)

Time taken by the taxi to reach the maximum speed is given as

`v_f - v_i = at`

`27 - 0 = 2(t)`

`t = 13.5 s`

now it moves with constant speed for next 41 s and then finally comes to rest in next 5 s

so total time for which it will move is given as

`T = 13.5 s + 41 s + 5 s`

`T = 59.5 s`

Part b)

Distance covered by the taxi while it accelerate to its maximum speed is given as

`d_1 = (v_f + v_i)/(2) t`

`d_1 = (27 m/s + 0)/(2) (13.5)`

`d_1 = 182.25`

Now it moves for constant speed for next 41 s so distance moved is given as

`d_2 = (27)(41)`

`d_2 = 1107 m`

Finally it comes to rest in next 5 s so distance moved in next 5 s

`d_3 = (v_f + v_i)/(2) t`

`d_3 = (27 + 0)/(2)(5)`

`d_3 = 67.5 m`

so here we have total distance moved by it is given as

`d = 182.25 + 1107 + 67.5`

`d = 1356.75 m`

average speed is given as

`v = (1356.75)/(59.5)`

`v = 22.8 m/s`