Answer to Question #120946 in Mechanics | Relativity for Vidurjah Perananthan

Question

In traffic knowledge, one speaks of reaction distance, braking distance and stopping distance. From the driver noticing the danger until they react, it takes a certain amount of reaction time. The reaction distance is the distance the car travels during the reaction time. The braking distance is the distance the car moves during braking. The stop distance is the sum of the reaction distance and the brake distance. Calculate the stopping distance for a motorist who has a speed of 108 km / h. The reaction time is assumed to be 0.84 s and the mean value of the acceleration during braking is 5.1 m / s2. Draw three diagrams for the whole process.
One with distance time-graph, one with speed time-graph and one with acceleration time-graph

Answer 1

Explanations & Calculations

• 108kmh^-1 = 30ms^-1

• Reaction distance is (d_r) : apply s = ut along the moving direction

$\qquad\qquad \begin{aligned} \small d_r &= \small 30ms^{-1} \times 0.84s =\bold{ 25.2m} \end{aligned}$

• Assuming the deceleration is uniform throughout, apply V = U +at to find the time taken until come to rest.

$\qquad\qquad \begin{aligned} \small 0ms^{-1} &= \small 30 ms^{-1} +(-5.1ms^{-2}) \times t\\ \small t &= \small \bold{5.88 s} \end{aligned}$

• Traveled distance after applying brakes (d₁) : apply V² = U² +2as

$\qquad\qquad \begin{aligned} \small 0 &= \small 30^2 +2\times(-5.1) \times d_1\\ \small d_1 &= \small \bold{88.24m }\end{aligned}$

• Stopping distance = 25.2 + 88.24 = 113.44 m
• Total time expelled = 0.84 + 5.88 = 6.72 s