A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelerate at 8.0 m/s^2. The two vehicles start the race at the same time and accelerate from rest. After 5.0 s, how fast is the sports car going? After 6.0 s, how distance will the motorcycle have gone? To make the race fair, the sports car starts 50.0 m ahead of the motorcycle. If the course is 200.0 m long, which vehicle wins the race?

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

Vf₁ = 25 m/s

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

s₂ = 90 m

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.