Question

An unmarked police car traveling a constant 95.0 km/h is passed by a speeder traveling 110 km/h . Precisely 2.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.00 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Answer 1

Answer: t = 7.61s

Explanation: The initial speed of the police's car is Vp = 95 km/h.

The initial speed of the car is Vc = 110km/h

The acceleration of the police's car is 2m/s^2

Now, we should write this the quantities in the same units, so lets write the velocities in meters per second.

1 kilometers has 1000 meters, and one hour has 3600 seconds, so we have that:

Vp = 95*1000/3600 m/s = 26.39m/s

Vc = 110*1000/3600 m/s = 30.56m/s

now, after the police car starts to accelerate, the velocity equation will be now.

The positions of the cars are:

P(t) = (a/2)*t^2 + v0*t + p0

Where a is the acceleration, v0 is the initial velocity, and p0 is the initial position.

We know that the difference in the velocity of the cars is two seconds, so after those the speeding car is:

(30.56m/s - 26.39m/s)*2s = 8.34m/s

Now we can write the position equations as:

Pp = 1m/s*t^2 + 26.39m/s*t + 0

Here i assume that the initial position of the car is at the 0 units in one axis.

Pc = 30.56m/s*t + 8.34m/s

Now we want to find the time at wich both positions are the same, and after that time the police car will go ahead of the speeding car.

1m/s*t^2 + 26.39m/s*t  = 30.56m/s*t + 8.34m/s

t^2 + (26.39 - 30.56)*t - 8.34 = 0

t^2 - 4.15*t - 8.34 = 0

Now we need to solve this quadratic equation:

t = (4.15 +/- √(4.15^2 - 4*1*(-8.34))/2 = (4.15 +/- 7.11)/2

From here we have two solutions, one positive and one negative, and we need to take the positive one:

t = (4.15 + 7.11)/2s = 11.26/2 s= 5.61s

And remember that the police car started accelerating two secnods after that the speeding car passed it, so the actual time is:

t = 5.61s + 2s = 7.61s

Answer 2

Answer:

So police car will overtake the speeder after 5.64 s

Explanation:

Initially the distance between police car and the speeder when police car is about to accelerate

$d = (v_1 - v_2)t$

$v_1 = 110 km/h = 110 \times \frac{1000}{3600}m/s = 30.55 m/s$

$v_2 = 95 km/h = 95 \times \frac{1000}{3600}m/s = 26.39 m/s$

$d = (30.55-26.39)(2) = 8.32 m$

now we have

now velocity of police with respect to speeder is given as

$v_r = v_2 - v_1 = 26.39 - 30.55 = -4.17 m/s$

relative acceleration of police car with respect to speeder

$a_r = a = 2 m/s^2$

now the time taken to cover the distance between police car and speeder is given as

$d = v_i t + \frac{1}{2}at^2$

$8.32 = -4.17 t + \frac{1}{2}(2)(t^2)$

$t^2 -4.17 t - 8.32 = 0$

$t = 5.64 s$