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

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

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

$d = \frac{v_i + v_f}{2} (t)$

$d = \frac{v_o + v_1}{2}(80)$

$1000 = 40(v_o + v_1)$

also we know that

$v_1 - v_o = a(80)$

also we have

$1000 = \frac{v_1 + v_2}{2}(120)$

$1000 = 60(v_1 + v_2)$

now we can say

$(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}$

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

$v_2 = v_o + at$

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

$v_2 = 5.77 m/s$