Question

Please answer the question

Answer 1

Reaction time = 0.6 seconds

Acceleration = -5 m/s²

Initial velocity = 30 kmph = 30×5/18 = 150/18 m/s

Final Velocity = 0 m/s

Using 2nd equation of uniformly accelerated motion.

$v = u + at \\ 0 = \frac{150}{8} + ( - 5 \times t) \\ - \frac{150}{8} = - 5t \\ 5t = \frac{150}{8} \\ t = \frac{150}{8 \times 5} \\ t = \frac{30}{18} \\ t = 1.6(approx..)$
without using reaction time the car will stop in 1.666 seconds.

but the driver reacted for 0.6 seconds.

so, time will be 1.6-0.6 = 1.0 seconds

To find distance, we can use 2nd equation of uniformly accelerated motion.

$s = ut + \frac{1}{2} a {t}^{2} \\ s = (\frac{150}{18} \times 1) + ( \frac{1}{2} \times - 5 \times 1) \\ s = \frac{150}{18} + \frac{ - 5}{2} \\ s = \frac{150 - 45}{18} \\ s = \frac{105}{18} \\ s = 5.833 \: metres$
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