Question

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.9 m when its initial speed is 80.0 km/h, and 25.7 m when its initial speed is 50.7 km/h. What are (a) your reaction time and (b) the magnitude of the deceleration?

Answer 1

Answer:

a) $t_r = 0.55 s$

b) a = 5.59 m/s²

Explanation:

given,

total distance traveled by the car to stop is 56.9 m when speed of vehicle is 80 km/h or 80 × 0.278 = 22.24 m/s

total distance traveled by the car to stop is 25.7 m when speed of vehicle is 50.7 km/h or 50.7 × 0.278 = 14.09 m/s

using stopping distance formula

$s_1 = v_1 t_r +\dfrac{v_1^2}{2 a}$................(1)

$s_2 = v_2 t_r +\dfrac{v_2^2}{2 a}$..............(2)

on solving both the equation we get

$a = \dfarc{v_1v_2(v_1-v_2)}{2(s_1v_2-s_2v_1)}$

$a = \dfarc{22.4\times 14.09(22.24-14.09)}{2(56.9\times 14.09-25.7\times 22.24)}$

a = 5.59 m/s²

now reaction time calculation

$t_r =\dfrac{v_1^2d_2-v_2^2d_1}{v_1v_2(v_1-v_2)}$

$t_r =\dfrac{22.24^2\times 25.7-14.09^2\times 56.9}{22.4\times 14.09(22.24-14.09)}$

$t_r = 0.55 s$