Question

A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the next 1 ????m=1000 m in 120 sec⁡. Calculate (a) the acceleration of the motorcycle and (b) its velocity at the beginning and end of the 2 km trip.

Answer 1

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Step-by-step explanation:

Part a)

Let the initial velocity of the motorcycle is

`v_i = v_o`

now at the end of 80 s let the speed is

`v_f = v_1`

after another 120 s let the speed will be

`v_f' = v_2`

now we know that

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

`d = (v_o + v_1)/(2)(80)`

`1000 = 40(v_o + v_1)`

also we know that

`v_1 - v_o = a(80)`

also we have

`1000 = (v_1 + v_2)/(2)(120)`

`1000 = 60(v_1 + v_2)`

now we can say

`(v_2 + v_1) - (v_o + v_1) = (50)/(3) - (50)/(2)`

also we know

`v_2 - v_o = a(120 + 80)`

`-8.33 = 200 a`

`a = -0.042 m/s^2`

Part b)

now we have

`v_1 + v_o = 25`

`v_1 - v_o = (-0.042)(80)`

`v_1 = 10.83 m/s`

so the starting velocity of the trip is

`v_o = 25 - 10.83 = 14.17 m/s`

now speed after t = 200 s is given as

`v_2 = v_o + at`

`v_2 = 14.17 - (0.042)(200)`

`v_2 = 5.77 m/s`