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 21.5 m/s , and the distance between them is 60.0 m . After t1 = 5.00 s , the motorcycle starts to accelerate at a rate of 8.00 m/s2 . The motorcycle catches up with the car at some time t2.

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.

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)? Should you need to use an answer from a previous part, make sure you use the unrounded value.

Answer 1

given data is A motorcycle is following a car that is traveling at constant speed on a straight highway.

• car and the motorcycle are both traveling at the same speed of 21.5 m/s
• distance between them is 60.0 m
• After t1 = 5.00 s , the motorcycle starts to accelerate at a rate of 8.00 m/s² .

Let

M = distance travelled by the motorcycle when it catches up with the car

M = 21.5(T₂) + (1/2)(a)(T₂)²

M = 21.5(T₂) + (1/2)(8)(T₂)²

M = 21.5(T₂) + 4(T₂)²

Let

C = distance travelled by the car when the motorcycle catches up with it

C = 21.5(T₂)

When the motorcycle catches up with the car, then

M = 60 + C

Therefore,

21.5(T₂) + 4(T₂)² = 60 + 21.5(T₂)

Since "21.5(T₂)" appears on both sides of the equation, it will simply cancel out, hence the above equation is modified to

4(T₂)² = 60

(T₂)² = 60/4=15

T₂ = 3.87 sec.

The motorcycle will catch up with the car 3.87 seconds after it has accelerated.

To determine the distance the motorcycle travelled from the time it accelerated,

M = 21.5(T₂) + 4(T₂)²

M = 21.5(3.87) + 4(3.87)²

M = 83.205 + 59.90

M = 143.105 m