Question

A car moves on a straight road. As the driver passes a point A on the road with a speed of 20 ms–1, he

notices an accident ahead at a point B. He immediately applies the brakes and the car moves with an
acceleration of ams–2, where a = – 6 and t s is the time after passing A. When t = 4, the car passes
the accident at B. The car then moves with a constant acceleration of 2 ms–2 until the original speed of
20 ms–1 is regained at a point C. Find
(i) the speed of the car at B, [4]
(ii) the distance AB, [3]
(iii) the time taken for the car to travel from B to C. [2]
Sketch the velocity-time graph for the journey from A to C.

Answer 1

The answer to the problem is:

(i) The speed of the car at B is -4 ms$^{-1}$.

(ii) The distance AB is 32 meters.

(iii) The time taken for the car to travel from B to C is 10 seconds.

Explanation:

To find the speed of the car at B, we can use the formula $v = v_0 + at$, where $v$ is the final speed, $v_0$ is the initial speed, $a$ is the acceleration, and $t$ is the time. In this case, $v_0 = 20$ ms$^{-1}$, $a = -6$ ms$^{-2}$, and $t = 4$ s. Plugging these values into the formula gives us:

$v = 20 + (-6)(4) = 20 - 24 = -4$ ms$^{-1}$

Thus, the speed of the car at B is -4 ms$^{-1}$.

To find the distance AB, we can use the formula $d = v_0t + \frac{1}{2}at^2$, where $d$ is the distance, $v_0$ is the initial speed, $a$ is the acceleration, and $t$ is the time. In this case, $v_0 = 20$ ms$^{-1}$, $a = -6$ ms$^{-2}$, and $t = 4$ s. Plugging these values into the formula gives us:

$d = (20)(4) + \frac{1}{2}(-6)(4^2) = 80 - 48 = 32$ meters

Thus, the distance AB is 32 meters.

To find the time taken for the car to travel from B to C, we can use the formula $v^2 = v_0^2 + 2ad$, where $v$ is the final speed, $v_0$ is the initial speed, $a$ is the acceleration, and $d$ is the distance. In this case, $v_0 = -4$ ms$^{-1}$, $v = 20$ ms$^{-1}$, $a = 2$ ms$^{-2}$, and $d$ is the distance from B to C. We can find this distance by using the formula for distance and setting $t = 0$, since we're looking for the time it takes the car to travel from B to C, not from A to B:

$d = (20)(0) + \frac{1}{2}(-6)(0^2) = 0 - 0 = 0$ meters

Thus, the distance from B to C is 0 meters. Plugging these values into the formula gives us:

$20^2 = (-4)^2 + 2(2)(0) = 16 + 0 = 16$

We can solve this equation for $t$ to find the time it takes the car to travel from B to C:

$t = \sqrt{\frac{v^2 - v_0^2}{2a}} = \sqrt{\frac{20^2 - (-4)^2}{2(2)}} = \sqrt{\frac{400}{4}} = \sqrt{100} = 10$ seconds

Thus, the time taken for the car to travel from B to C is 10 seconds.

The velocity-time graph for the journey from A to C would have a positive slope from A to B, a negative slope from B to C, and a positive slope from C to the end of the journey.