Question

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 23.0 m/s , and the distance between them is 53.0 m . After t1 = 4.00 s , the motorcycle starts to accelerate at a rate of 7.00 m/s2 . The motorcycle catches up with the car at some time t2.

Answer 1

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car  = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

now, equating both the distances

$23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

$t_2^2 = 15.143$

    t₂ = 3.89 s

time taken by the motorcycle to catch the car is equal to 3.89 s