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 19.0m/s , and the distance between them is 52.0m . After t1 = 5.00s , the motorcycle starts to accelerate at a rate of 5.00m/s2. 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
Please explain
your basic equation of motion is
s = s0 + vt + 1/2 at^2
t2 = t1+5
19.0t1 + 5.00/2 t1^2 = 52 + 19.0t2
Thanks man
To find the time it takes from when the motorcycle starts to accelerate until it catches up with the car, we need to determine the time it takes for the motorcycle to cover the initial distance between them and then surpass the car.
First, let's find the initial distance the motorcycle travels before it starts to accelerate.
The initial speed of the motorcycle is the same as the car's speed, which is 19.0 m/s. So, in 5.00 seconds, the motorcycle would have covered a distance of:
Distance = Speed x Time
Distance = 19.0 m/s x 5.00 s
Distance = 95.0 m
Now, let's calculate how long it takes for the motorcycle to catch up with the car.
To catch up with the car, the motorcycle needs to cover the remaining distance between them, which is the initial distance (52.0 m) minus the distance it already traveled (95.0 m).
Remaining Distance = Initial Distance - Distance Already Traveled
Remaining Distance = 52.0 m - 95.0 m
Remaining Distance = -43.0 m
Since the remaining distance is negative, it means the motorcycle has already surpassed the car when it starts to accelerate. So, the time it takes for the motorcycle to catch up with the car after it starts to accelerate is 0 seconds (t2 - t1 = t2 - 5.00 = 0).
In other words, the motorcycle catches up with the car immediately after it starts to accelerate.