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

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $9$
$u = 10 m / s, v = 20 m / s, x = 130 m$
$a = \frac{v - u}{t} = \frac{20 - 10}{t} = \frac{10}{t}$
$s = u t + \frac{1}{2} a t^{2} ⟹ 135 = 10 t + \frac{1}{2} \frac{10}{t} t^{2} ⟹ t = 9 s$

Suggest Corrections

Similar questions

A particle is moving in a straight line with a constant acceleration. It changes its velocity from 10 $\frac{m}{s}$ to 20 $\frac{m}{s}$ in ‘t’ second. If the displacement of the particle during this interval is 135 m, the value of ‘t’ is

Q. A particle moving in a straight line with initial velocity

u

and uniform acceleration

a

. If the sum of the distance travelled in

t^{t h}

and

{(t + 1)}^{t h}

seconds is

100 c m

, then its velocity after

t

seconds in

c m / s

A particle starts to move in a straight line from a point with velocity

10 m s^{- 1}

and acceleration

- 2.0 m s^{- 2}

. Find position and velocity of the particle at t = 10 s

Join BYJU'S Learning Program

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 135 min t second. The value of t is

The correct option is C 9u=10m/s,v=20m/s,x=130ma=v−ut=20−10t=10ts=ut+12at2⟹135=10t+1210tt2⟹t=9s

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

The correct option is C $9$
$u = 10 m / s, v = 20 m / s, x = 130 m$
$a = \frac{v - u}{t} = \frac{20 - 10}{t} = \frac{10}{t}$
$s = u t + \frac{1}{2} a t^{2} ⟹ 135 = 10 t + \frac{1}{2} \frac{10}{t} t^{2} ⟹ t = 9 s$