a police car is cruising on the right lane at 25 m/s (meters/sec.). A motorcycle, on the left lane, passes the police car at a constant 35 m/s. The police officer reacts fast and accelerates at a constant 5 m/s^2.

(a)When does he catch up?
(b)What was his velocity at that instant?

At the time t after the police car is passed, the distances travelled are:

X1 (motorcycle) = 35 t (meters)
X2 (police) = 25t + (1/2)(5)t^2

(a) Set X1 = X2 and solve for t. Ignore the t=0 solution. Take the other one.

(b) V2 = 25 + 5t is the celocity of the police car. Use the appropriate value of t.

thank you!

erft

To find the time when the police car catches up to the motorcycle and the velocity of the police car at that instant, we can follow these steps:

Step 1: Determine the relative velocity between the police car and the motorcycle.
The relative velocity is the difference in velocity between the two vehicles. In this case, the motorcycle is moving at a constant velocity of 35 m/s, and the police car is accelerating at a constant rate. Therefore, the relative velocity will be the difference between their velocities: 35 m/s - 25 m/s = 10 m/s.

Step 2: Calculate the time it takes for the police car to catch up to the motorcycle.
Since the police car is accelerating, it will gradually reduce the gap between them until it catches up. We can use the formula:
Time (t) = Relative distance / Relative velocity
In this case, the relative distance is initially unknown, but we can use the equation for uniform acceleration to find it:
d = u*t + 0.5*a*t^2
where d is the relative distance, u is the initial velocity (25 m/s), t is the time, and a is the acceleration (5 m/s^2).

Since the motorcycle's velocity is constant, the relative distance can be expressed as:
d = (35 m/s) * t

So, combining the two equations, we have:
(35 m/s) * t = (25 m/s) * t + 0.5 * (5 m/s^2) * t^2
35t = 25t + 2.5t^2

Step 3: Solve the equation to find the time.
Rearranging the equation, we get:
2.5t^2 - 10t = 0
Dividing both sides by 2.5, we have:
t^2 - 4t = 0
Factorizing, we get:
t(t - 4) = 0
So, t = 0 or t = 4

Since time cannot be zero in this context, we discard t = 0. Therefore, the time it takes for the police car to catch up to the motorcycle is 4 seconds.

Step 4: Calculate the velocity of the police car at that instant.
We can use the formula for uniform acceleration to find the velocity at that instant:
v = u + a * t
where v is the final velocity (which is the velocity of the police car at the time of catching up), u is the initial velocity (25 m/s), a is the acceleration (5 m/s^2), and t is the time (4 seconds).

Substituting the values, we get:
v = 25 m/s + (5 m/s^2) * 4 s
v = 25 m/s + 20 m/s
v = 45 m/s

So, the velocity of the police car at the instant of catching up to the motorcycle is 45 m/s.