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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Rest and Motion Kinematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rest and Motion Kinematics

A driver takes 0.20 s to apply the brakes after seeing a need for...

A driver takes 0.20 s0.20 \text{ s} to apply the brakes after seeing a need for it. This is termed the reaction time of the driver. If he is driving a car at a speed of 54 km/h54 \text{ km/h} and the brakes cause a deceleration of 6.0 m/s26.0 \text{ m/s}^2, determine the distance travelled by the car after he sees the need to put the brakes on, in m.

Options

  1. A18.7518.75
  2. B21.7521.75
  3. C3.003.00
  4. D253.80253.80

Correct answer

B. 21.7521.75

Step-by-step solution

Initial speed of the car, u=54 km/h=54×518 m/s=15 m/su = 54 \text{ km/h} = 54 \times \dfrac{5}{18} \text{ m/s} = 15 \text{ m/s}

Reaction time of the driver, t=0.20 st = 0.20 \text{ s}

Distance travelled during the reaction time, s1=u×t=15×0.20=3.0 ms_1 = u \times t = 15 \times 0.20 = 3.0 \text{ m}

Deceleration of the car, a=−6.0 m/s2a = -6.0 \text{ m/s}^2

Final velocity, v=0v = 0

Using the equation of motion, v2−u2=2as2v^2 - u^2 = 2as_2

0−(15)2=2×(−6.0)×s20 - (15)^2 = 2 \times (-6.0) \times s_2

s2=22512=18.75 ms_2 = \dfrac{225}{12} = 18.75 \text{ m}

Total distance travelled by the car, s=s1+s2=3.0+18.75=21.75 ms = s_1 + s_2 = 3.0 + 18.75 = 21.75 \text{ m}
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