A student driving at a constant speed of 45 m/sec passes a highway patrol car parked on the side of the road. One second later, the patrol car accelerates at a constant rate of 3.00 m/sec^2 in an attempt to catch up with the speeding student. How long does it take to over take the student?

Question

A student driving at a constant speed of 45 m/sec passes a highway patrol car parked on the side of the road. One second later, the patrol car accelerates at a constant rate of 3.00 m/sec^2 in an attempt to catch up with the speeding student. How long does it take to over take the student?

Answer 1

After one second, when the patrol car is at location x = 0, the student is at 45 m.

The student's location as a function of time, where t=0 when the police car starts accelerating is

45 + 45*t

The policeman's position is

1/2*3*t^2

Set these two equations equal to each other and solve for t

Answer 2

To find the time it takes for the patrol car to overtake the student, we can use a simple equation of motion.

Let's assume that the time it takes for the patrol car to catch up with the student is 't' seconds.

For the student's motion, we know that the speed is constant at 45 m/s (as given). Therefore, we can say that the distance covered by the student in 't' seconds is equal to 45t meters.

For the patrol car's motion, it starts from rest (initial speed is 0 m/s). It then accelerates at a constant rate of 3.00 m/s^2. So, we can use the equation of motion:

distance = initial velocity * time + (1/2) * acceleration * time^2

The initial velocity of the patrol car is 0 m/s, the acceleration is 3.00 m/s^2, and the time is (t + 1) seconds because it starts 1 second later than the student.

Therefore, the distance covered by the patrol car in (t + 1) seconds is (1/2) * 3.00 * (t + 1)^2 meters.

Since the patrol car catches up with the student, the distances covered by both the student and the patrol car are the same.

Therefore, we can set up the following equation:

45t = (1/2) * 3.00 * (t + 1)^2

Simplifying this equation, we get:

45t = 1.5 * (t + 1)^2

Now, we can solve this equation to find the value of 't' which represents the time it takes for the patrol car to overtake the student.

So, let's expand the equation:

45t = 1.5 * (t^2 + 2t + 1)

45t = 1.5t^2 + 3t + 1.5

Rearranging the equation, we get:

1.5t^2 + 3t + 1.5 - 45t = 0

1.5t^2 - 42t + 1.5 = 0

Now, solving this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

a = 1.5, b = -42, c = 1.5

Plugging in the values, we can calculate 't'.