SATHEE: Kinematics Question 160

Kinematics Question 160

Question: A particle moves with uniform acceleration along a straight line AB. it’s velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes $t_{1}$ seconds to go from A to M and $t_{2}$ seconds to go from M to B. Then $t_{2} / t_{1}$ it’s

Options:

A) 1 : 1

B) 2 : 1

C) 1:2

D) 3:1

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $V^{2} = 2^{2} + \frac{2 a s}{2}$
$\Rightarrow v^{2} - 4 = a s$

$14^{2} - v^{2} = \frac{2 a s}{2}$
$\Rightarrow 196 - v^{2} = a s$ From (i) and (ii) $v^{2} - 4 = 196 - v^{2} \Rightarrow v = 10 m / s$ Now $t_{1} = \frac{v - 2}{a} = \frac{10 - 2}{a} = \frac{8}{a}$

$t_{2} = \frac{14 - v}{a} = \frac{14 - 10}{a} = \frac{4}{a} \Rightarrow \frac{t_{2}}{t^{1}} = \frac{4 / a}{8 / a} = \frac{1}{2}$

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Kinematics Question 160

Question: A particle moves with uniform acceleration along a straight line AB. it’s velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes t1 seconds to go from A to M and t2 seconds to go from M to B. Then t2/t1 it’s

Options:

Answer:

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Question: A particle moves with uniform acceleration along a straight line AB. it’s velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes $t_{1}$ seconds to go from A to M and $t_{2}$ seconds to go from M to B. Then $t_{2} / t_{1}$ it’s