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

Question

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

Tardigrade Admin · Accepted Answer

v2-u2=2 a s Given v=20 ms -1, u=10 ms -1, s=135 m ∴ a=(400-100/2 × 135)=(300/270) =(10/9) m / s 2 v=u+a t ⇒ t=(v-u/a)=(10 m / s /(10)9 m / s 2) =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 $t$ second. The value of $t$ is

10

1.8

12

9

$v^{2} - u^{2} = 2 a s$
Given $v = 20 m s^{- 1},$
$u = 10 m s^{- 1},$
$s = 135 m$
$∴ a = \frac{400 - 100}{2 \times 135} = \frac{300}{270}$
$= \frac{10}{9} m / s^{2}$
$v = u + a t$
$\Rightarrow t = \frac{v - u}{a} = \frac{10 m / s}{\frac{10}{9} m / s ^{2}}$
$= 9 s$

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 second. The value of t is

10

1.8

12

9

v2−u2=2as Given v=20ms−1, u=10ms−1, s=135m ∴a=2×135400−100​=270300​ =910​m/s2 v=u+at ⇒t=av−u​=910​m/s210m/s​ =9s

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 $t$ second. The value of $t$ is

$v^{2} - u^{2} = 2 a s$
Given $v = 20 m s^{- 1},$
$u = 10 m s^{- 1},$
$s = 135 m$
$∴ a = \frac{400 - 100}{2 \times 135} = \frac{300}{270}$
$= \frac{10}{9} m / s^{2}$
$v = u + a t$
$\Rightarrow t = \frac{v - u}{a} = \frac{10 m / s}{\frac{10}{9} m / s ^{2}}$
$= 9 s$