A man driving his car at 28 metres per second saw a child crossing the road in front of him. Given that the man’s response time was 0.6 seconds and that after he struck the brakes, the car decelerated uniformly at a rate of 10 metres per second squared until it stopped, find the total stopping distance of the car.
We can calculate the total distance by adding the response distance and the braking distance. Each of these can be calculated using our equations of motion or SUVAT equations. During the response time or response distance, the acceleration is equal to zero as the speed is constant — 28 metres per second. This gives us that 𝑢 is equal to 28, 𝑣 is also equal to 28, 𝑎 is equal to zero, 𝑡 is equal to 0.6, and 𝑠 is the unknown.
In order to calculate 𝑠, we’ll use the equation of motion: 𝑢 plus 𝑣 divided by two multiplied by 𝑡. Substituting in our values gives us 28 plus 28 divided by two multiplied by 0.6. This is equal to 16.8. Therefore, our response distance is 16.8 metres.
To calculate the braking distance, we know that the initial velocity is equal to 28, the final velocity was equal to zero as the car came to stop, 𝑎 is equal to negative 10 as the car was decelerating, 𝑠 once again is our unknown. This time we’re going to use the equation 𝑣 squared equals 𝑢 squared plus two 𝑎 𝑠.
Substituting in our values of 𝑢, 𝑣, and 𝑎 gives us zero is equal to 784 minus 20𝑠. This can be rewritten as 20𝑠 is equal to 784. And dividing both sides by 20 gives us a value of 𝑠 of 39.2. Therefore, the braking distance is 39.2 metres.
The total stopping distance is then equal to 16.8 plus 39.2, which is equal to 56 metres. In practical terms, this means the man is able to stop the car 56 metres after he first sees the child.