A motorcycle is following a car that is traveling at constant speed on a straigh

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 47.0 {mph}, 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/s^2}. The motorcycle catches up with the car at some time t2.

(A) How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1.

(B) How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?

Explanation / Answer

Given Initial speed of motor cycle and car     v = 47 m/p                                                                = 47*0.44704 m/s                                                                = 21 m/s   Distance between car and moter cycle d = 53 m By using kinematioc equations                 x = x_o + v_o t + (1/2) a t2       For car                   x_c   = x_i + v t     since the acceleration of the car is zero       For motor cycle                x_m   = vt + ( 1/2) a t 2         x_c = x_m is the distance displaced when they meet each other    there fore               x_i + v t   = vt + ( 1/2) a t 2                       x_i   = ( 1/2) a t^2       where acceleration of motor cycle is   a = 7m/s       on substituting the numerical values                    53 m = ( 1/2) ( 7 ) t ^2         there timedoes it take from the moment when the motorcycle starts to accelerate until it catches up with the car                               t =   3.89m/s         a)   t2 - t1     =  4 - 3.89   = 0.1086 s          b)                   s = v t + ( 1/2) a t2                     = ( 21 ) ( 3.89 ) + ( 1/2 ) ( 7)(3.89 ) ^2                   =   134.6523 m   

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