A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m s- 1 to 20 m s- 1 while passing through a distance 135 m in t seconds. The value of t is:

Question

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m s- 1 to 20 m s- 1 while passing through a distance 135 m in t seconds. The value of t is: (A) 10 (B) 1.8 (C) 12 (D) 9

Tardigrade Admin · Accepted Answer

Initial velocity, u = 10 m s- 1 Final velocity , v = 20 m s- 1 Distance , s = 135 m Let , acceleration = a Using the formula , v2 = u2 + 2 a s ⇒ a = (v2 - u2/2 s) or a = ((. 20 .)2 - (. 10 .)2/2 × 135) = (400 - 100/2 × 135) or a = (300/2 × 135) = (150/135) = (10/9) m s- 2 Now , using the relation , v = u + a t

t = \frac{v â u}{a} = \frac{20 â 10}{\frac{10}{9}} = \frac{10}{10} Ã 9 = 9 s .

Q. A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 m s^{- 1}$ to $20 m s^{- 1}$ while passing through a distance $135 m$ in $t$ seconds. The value of $t$ is:

$10$

$1.8$

$12$

$9$

Q. A particle moves in a straight line with a constant acceleration. It changes its velocity from 10ms−1 to 20ms−1 while passing through a distance 135m in t seconds. The value of t is:

10

1.8

12

9

Q. A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 m s^{- 1}$ to $20 m s^{- 1}$ while passing through a distance $135 m$ in $t$ seconds. The value of $t$ is:

$10$

$1.8$

$12$

$9$