Answer to Question #216604 in Math for Kwame

Question

A car spends through a police checkpoint at 15ms and Continues on strought rond out the same spoil. Three seconds later, a police motor cyclist at the check point starting from rest, accelerates uniformly in persuit of the car. In another 15 seconds the policeman, then travelling at Vers" and abst with the car, Signals to the driver to skup. The police an himself decelerates uniformly and stique If seconds from west instant. The car, however, trounds on at the same britial sped of 15ms for another 2 Seconds before the brakes are applied bringing the car to stop under uniform recordation in Highlands Calculate the value of V. this the distance between the car and the motor cylist after but have stopped

Answer 1

1. Let $a_1=$ the constant acceleration when a police motorcyclist accelerates uniformly in pursuit of the car.

$$V=0+a_1(15)$$

$$a_1=\dfrac{V}{15}$$

The distance travelled by the car to the meeting is

$$d_{car1}=15\ \dfrac{m}{s}\ (3s+15 s)=270\ m$$

The distance travelled by the a police motorcyclist to the meeting is

$$d_{motor1}=0+0+\dfrac{a_1(15)^2}{2}=\dfrac{225}{2}a_1=\dfrac{15}{2}V$$

Then

$$\dfrac{15}{2}V=270$$

$$V=36\ \dfrac{m}{s}$$

Let $a_2=$ the constant acceleration when a police motorcyclist decelerates uniformly to stop.

$$0-V=a_2(4)$$

$$a_2=-\dfrac{36\ \dfrac{m}{s}}{4s}=-9\ \dfrac{m}{s^2}$$

The distance covered by the police motorcyclist to stop is

$$d_2=0+V(4)+\dfrac{a_2(4)^2}{2}$$

$$d_2=0+36\ \dfrac{m}{s}(4s)+\dfrac{-9\ \dfrac{m}{s^2}(4s)^2}{2}=72\ m$$

Let $a_3=$ the constant acceleration when a car decelerates uniformly to stop.

$$0-15=a_3(4)$$

$$a_3=-\dfrac{15\ \dfrac{m}{s}}{4s}=-3.75\ \dfrac{m}{s^2}$$

The distance covered by the police motorcyclist to stop is

$$d_3=15(2)+15(4)+\dfrac{a_3(4)^2}{2}$$

$$d_3=15\ \dfrac{m}{s}(2s)+15\ \dfrac{m}{s}(4s)+\dfrac{-3.75\ \dfrac{m}{s^2}(4s)^2}{2}=60\ m$$

The distance between the car and the motorcyclist after both have stopped is

$$72\ m-60\ m=12\ m$$