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Motion in A Straight Line question

2021 · 26 Feb · Shift 2 · Q51
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  5. /2021 · 26 Feb · Shift 2 · Q51

Motion in A Straight Line question

2021 · 26 Feb · Shift 2 · Q51

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of t1t2{{{t_1}} \over {{t_2}}}t2​t1​​ wil be :
  1. A
    a1+a2a2{{{a_1} + {a_2}} \over {{a_2}}}a2​a1​+a2​​
  2. B
    a1+a2a1{{{a_1} + {a_2}} \over {{a_1}}}a1​a1​+a2​​
  3. C
    a2a1{{{a_2}} \over {{a_1}}}a1​a2​​
  4. D
    a1a2{{{a_1}} \over {{a_2}}}a2​a1​​
View written solutionFree

Correct answer: C

  1. Motion during acceleration

The scooter starts from rest and accelerates with constant acceleration a1a_1a1​ for time t1t_1t1​.

Using v=u+atv = u + atv=u+at we get v=0+a1t1=a1t1v = 0 + a_1 t_1 = a_1 t_1v=0+a1​t1​=a1​t1​

So, the velocity after the first phase is v=a1t1v = a_1 t_1v=a1​t1​

  1. Motion during retardation

Now the scooter slows down with constant retardation a2a_2a2​ for time t2t_2t2​ and finally comes to rest.

Here initial velocity for this phase is v=a1t1v = a_1 t_1v=a1​t1​, final velocity is 000, and acceleration is −a2-a_2−a2​.

Again using v=u+atv = u + atv=u+at 0=a1t1−a2t20 = a_1 t_1 - a_2 t_20=a1​t1​−a2​t2​

Thus, a1t1=a2t2a_1 t_1 = a_2 t_2a1​t1​=a2​t2​

  1. Find the ratio

t1t2=a2a1\frac{t_1}{t_2} = \frac{a_2}{a_1}t2​t1​​=a1​a2​​

  1. Match with options

Option C is a2a1\frac{a_2}{a_1}a1​a2​​ which matches our result.

Therefore, the correct answer is Option C.

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