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Motion in A Straight Line question

2018 · 15 Apr · Shift 1 · Q66
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Motion in A Straight Line question

2018 · 15 Apr · Shift 1 · Q66

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
An automobile, travelling at 40 40\,40 km/h, can be stopped at a distance of 40 40\,40 m by applying brakes. If the same automobile is travelling at 80 80\,80 km/h, the minimum stopping distance, in metres, is (assume no skidding) :
  1. A
    45 45\,45 m
  2. B
    100 100\,100 m
  3. C
    150 150\,150 m
  4. D
    160 160\,160 m
View written solutionFree

Correct answer: D

  1. Use the kinematic relation for stopping distance

When brakes are applied with the same braking force and there is no skidding, the retardation aaa is constant.

For stopping, v2=u2+2asv^2=u^2+2asv2=u2+2as Since final velocity v=0v=0v=0, 0=u2+2as0=u^2+2as0=u2+2as s=−u22as=-\frac{u^2}{2a}s=−2au2​

Thus, for the same automobile under the same braking conditions, s∝u2s \propto u^2s∝u2 That is, stopping distance is proportional to the square of speed.

  1. Compare the two cases

Initial speed changes from 404040 km/h to 808080 km/h.

So, u2u1=8040=2\frac{u_2}{u_1}=\frac{80}{40}=2u1​u2​​=4080​=2

Therefore, s2s1=(u2u1)2=22=4\frac{s_2}{s_1}=\left(\frac{u_2}{u_1}\right)^2=2^2=4s1​s2​​=(u1​u2​​)2=22=4

Given, s1=40 ms_1=40\text{ m}s1​=40 m So, s2=4×40=160 ms_2=4\times 40=160\text{ m}s2​=4×40=160 m

  1. Check options
  • A: 454545 m
  • B: 100100100 m
  • C: 150150150 m
  • D: 160160160 m

Hence, the correct option is D.

Final Answer

160 m160\text{ m}160 m

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