A particle moving along a straight line with constant acceleration of $- 2 m / s^{2}$ has a velocity of $6 m / s$ at $t = 0$ sec. Find the ratio of distance and displacement at $t = 5$ sec.

13 : 5

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17 : 8

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9 : 5

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11 : 3

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Solution

The correct option is A 13 : 5
$v = u + a t = 6 - 2 t$
$v = 6 - 2 t$
If $v = 0 \Rightarrow 6 - 2 t = 0 \Rightarrow 2 t = 6 \Rightarrow t = 3 s e c$
At $t = 5 \Rightarrow v = 6 - 2 (5) = - 4 m / s^{2}$

Velocity time graph for given situation can be considered as

Distance $= A_{1} + A_{2} = \frac{1}{2} (6) (3) + \frac{1}{2} (2) (4) = 13$
Displacement $= A_{1} - A_{2} = \frac{1}{2} (6) (3) - \frac{1}{2} (2) (4) = 5$
Threfore, the ratio $= 13 : 5$

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A particle moving along a straight line with constant acceleration of −2 m/s2 has a velocity of 6 m/s at t=0 sec. Find the ratio of distance and displacement at t=5 sec.

The correct option is A 13 : 5v=u+at=6−2t v=6−2t If v=0⇒6−2t=0⇒2t=6⇒t=3 sec At t=5⇒v=6−2(5)=−4 m/s2 Velocity time graph for given situation can be considered as Distance =A1+A2=12(6)(3)+12(2)(4)=13 Displacement =A1−A2=12(6)(3)−12(2)(4)=5 Threfore, the ratio =13:5

A particle moving along a straight line with constant acceleration of $- 2 m / s^{2}$ has a velocity of $6 m / s$ at $t = 0$ sec. Find the ratio of distance and displacement at $t = 5$ sec.