A car is speeding at constant velocity of 30ms-1 in an 80kmh-1 zone as it passes a stationary police motorcycle hiding. The motorcycle starts to accelerate at 5ms-2 at the same instant the car passes it and continues for 8 seconds. It then stops accelerating and maintains its speed until it catches the car.

a) How long does it take for the motorcycle to catch up to the car?
b) How far did the motorcycle travel before catching up to the car?

Please explain how you got the answer

The distance the motorcycle and the car go is the same. So set up an equation for the distance the motorcycle goes and it will equal the car's accelerating distance and constant distance. You can use d=rt for the constant portions.

thanks :)

To solve this problem, we can break it down into two parts: the initial acceleration of the motorcycle and the subsequent constant speed of the motorcycle.

a) How long does it take for the motorcycle to catch up to the car?
1. Convert the speed limit of the car from kilometers per hour (km/h) to meters per second (m/s). We can do this by dividing by 3.6 since 1 km/h is equal to (1/3.6) m/s.
Speed limit = 80 km/h = (80/3.6) m/s = 22.22 m/s

2. Determine the initial distance between the motorcycle and the car at the instant when the motorcycle starts to accelerate.
To do this, we need to calculate the distance traveled by the car in the 8 seconds.
Distance car traveled = velocity × time = 30 m/s × 8 s = 240 m

3. Calculate the initial velocity of the motorcycle at the instant when it starts accelerating.
The initial velocity of the motorcycle is the same as the car's velocity since it is stationary when the car passes it.
Initial velocity of motorcycle = 30 m/s

4. Use the equation of motion for the motorcycle to find the time it takes for it to catch up to the car.
The equation for distance traveled during uniform acceleration is given by: s = ut + (1/2)at^2
where s = distance, u = initial velocity, a = acceleration, and t = time.

Since we want to find the time when the motorcycle catches up to the car, we set the distance traveled by the motorcycle equal to the initial distance between the car and motorcycle.
Therefore, 240 = (30 × t) + (1/2)(5)(t^2)

Simplify and rearrange the equation to solve for t:
240 = 30t + (5/2)t^2 => 5t^2 + 60t - 480 = 0

Solve this quadratic equation either by factoring, completing the square, or using the quadratic formula.
Either 5(t + 8)(t - 12) = 0 => t = -8 or t = 12
Since time cannot be negative in this context, the motorcycle catches up to the car after 12 seconds.

b) How far did the motorcycle travel before catching up to the car?
1. Using the time calculated in the previous step (t = 12 seconds), substitute it into the equation of motion for the motorcycle to find the distance traveled by the motorcycle during acceleration.
s = ut + (1/2)at^2
s = 30(12) + (1/2)(5)(12^2)
s = 360 + 360 = 720 meters

2. Determine the distance traveled by the motorcycle after it stops accelerating.
Since the motorcycle maintains a constant velocity after accelerating, the distance traveled can be calculated using the equation: distance = velocity × time.
Distance = 22.22 m/s × 12 s = 266.64 meters

3. The total distance traveled by the motorcycle before catching up to the car is the sum of the distances calculated above.
Total distance = distance during acceleration + distance after acceleration
Total distance = 720 meters + 266.64 meters = 986.64 meters

Therefore, the motorcycle takes 12 seconds to catch up to the car, and it travels 986.64 meters before catching up to the car.