A point initially at rest moves along $X - a x i s$ . Its acceleration varies with time as $a = (6 t + 5) m s^{- 2}$ . If it starts from origin, the distance covered in $2 s$ is

$20 m$

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

$18 m$

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

$16 m$

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

$25 m$

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

Open in App

Solution

The correct option is B
$18 m$

Step 1: Given data
Acceleration of a point, $a = (6 t + 5) m s^{- 2}$
Time taken by point, $t = 2 s$
Step 2: Find velocity, $v$
Velocity, $V = \int a d t$ [As, $a = \frac{d v}{d t}$ ]
$= \int (6 t + 5) d t = (6 / 2) t^{2} + 5 t$
$= 3 t^{2} + 5 t$
Step 3: Find distance, $s$
As we know that, $v = \frac{d s}{d t}$
Where, “s” is the displacement/ distance.
Distance, $d s = v d t$
$s = \int (3 t^{2} + 5 t) d t$ [As, $v = \frac{d s}{d t}$ ]
$= t^{3} + 5 t^{2} / 2$
At $t = 2 s$
Distance, $s = 8 + (5 x 4) / 2$
$= 18 m$
Hence, option (B) is correct.

Suggest Corrections

Similar questions

Question 15
A car starts from rest and moves along the x-axis with constant acceleration $5 m s^{- 2}$ for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Q. A point initially at rest moves along

x -

axis. Its acceleration varies with time as

a = (6 t + 5) {m / s}^{2}

. If it starts from origin, the distance covered in

2 s

is.

Q. A point mass initially at rest moves along

x -

axis. Its acceleration varies with time as

a = 6 t + 5 m / s^{2}

. If its starts from origin, the distance covered by it in

1 s

Q. A point initially at run moves along x-axis. Its acceleration varies with time

a = (6 t + 5)

in m/s. If its starts from origin, the distance covered in

2 s

is :

Q. A car starts from rest and moves along the x-axis with a constant acceleration

5 m s^{- 2}

for

8 s

. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?

Join BYJU'S Learning Program

A point initially at rest moves along X-axis. Its acceleration varies with time as a=(6t+5)ms-2. If it starts from origin, the distance covered in 2s is

A point initially at rest moves along $X - a x i s$ . Its acceleration varies with time as $a = (6 t + 5) m s^{- 2}$ . If it starts from origin, the distance covered in $2 s$ is