A car driver travelling at 72 h-l suddenly sees a fallen tree on the road 40 m ahead. Heputs on the brakes to stop before he hits the tree. To put on the brakes requires 0.75 s(reaction time of the driver), after which the retardation is 8 m s-2.a. What is the total stopping time?b. How far does he travels before the brakes are applied?c. What is his total stopping distance?d. If he subsequently travels at twice the speed, how far ahead should he be ableto see clearly for safety? (Assume the deceleration is the same.)
Question
A car driver travelling at 72 h-l suddenly sees a fallen tree on the road 40 m ahead. Heputs on the brakes to stop before he hits the tree. To put on the brakes requires 0.75 s(reaction time of the driver), after which the retardation is 8 m s-2.a. What is the total stopping time?b. How far does he travels before the brakes are applied?c. What is his total stopping distance?d. If he subsequently travels at twice the speed, how far ahead should he be ableto see clearly for safety? (Assume the deceleration is the same.)
Solution
a. Total stopping time:
First, we need to convert the speed from km/h to m/s. We do this by dividing by 3.6 (since 1 km = 1000 m and 1 hour = 3600 seconds). So, 72 km/h = 72/3.6 = 20 m/s.
The total stopping time is the sum of the reaction time and the time it takes for the car to stop after the brakes are applied. The time it takes for the car to stop can be calculated using the formula t = v/a, where v is the initial speed and a is the acceleration (in this case, deceleration, so we take it as -8 m/s^2).
So, t = 20/8 = 2.5 seconds.
Therefore, the total stopping time is 0.75 s (reaction time) + 2.5 s (braking time) = 3.25 seconds.
b. Distance travelled before brakes are applied:
This can be calculated using the formula d = vt, where v is the speed and t is the time.
So, d = 20 m/s * 0.75 s = 15 meters.
c. Total stopping distance:
The stopping distance after the brakes are applied can be calculated using the formula d = (v^2) / (2a), where v is the initial speed and a is the acceleration (deceleration in this case).
So, d = (20^2) / (2*8) = 25 meters.
Therefore, the total stopping distance is 15 m (distance travelled during reaction time) + 25 m (distance travelled during braking) = 40 meters.
d. Distance needed to stop if speed is doubled:
If the speed is doubled, the initial speed becomes 40 m/s. The stopping distance at this speed can be calculated using the same formula as above: d = (v^2) / (2a).
So, d = (40^2) / (2*8) = 100 meters.
However, the reaction time is still the same, so the car will travel 40 m/s * 0.75 s = 30 meters before the brakes are applied.
Therefore, the total distance needed to stop safely at this speed is 30 m (distance travelled during reaction time) + 100 m (distance travelled during braking) = 130 meters.
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