Amotorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. calculate (a) the acceleration of the motorcycle and (b) its velocity at the beginning and end of the 2-km trip.

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time  t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²    

put value

1000 = v(80) + 1/2×a(80)²           1

and

2000 = v(200) + 1/2×a(200)²      2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

so by kinematic final velocity will be

V² = v² + 2ad

V² = (14.167)² + 2×(-0.042)×(2000)

V²  = 32.70

V = 5.7183 m/s

so

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s