Answer to Question #216160 in Math for eyes

Question

: a car speed through a police checkpoint at a speed of 15m/s and continues in a straight road at the same speed. three seconds later, a police motorcyclist at the checkpoint starting from rest, accelerate uniformly in pursuit of the car. in another 15s the policeman than travelling at Vm/s and abreast with the car, signals the driver to stop. The policeman himself decelerates uniformly and stop 4s from that instant. The car however, travels on at the same initial speed at 15m/s for another 2s before the brakes are applied bringing the car to stop under uniform retardation in 4s. calculate the value of V. And the distance between the car and the motorcyclist after both 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

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