The displacement x of a body moving in a straight line is related to time t as $2 x^{2} + 3 x = t$ . If v is the velocity of the body at a certain instant of time, its acceleration will be

$- v^{3}$

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

$- 2 v^{3}$

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

$- 3 v^{3}$

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

$- 4 v^{3}$

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

Open in App

Solution

The correct option is D
$- 4 v^{3}$

Differentiating $2 x^{2} + 3 x = t$ with respect to t, $4 x \frac{d x}{d t} + \frac{3 d x}{d t} = 1$
Now, $\frac{d x}{d t} = v$
$∴ 4 x v + 3 v = 1$
$\Rightarrow 4 x + 3 = \frac{1}{v}$
Diffrentiating with respect to time t,
$4 \frac{d x}{d t} = \frac{- 1}{v^{2}} \frac{d v}{d t}$
$\Rightarrow 4 v = \frac{- 1}{v^{2}} \frac{d v}{d t}$
$\Rightarrow a = \frac{d v}{d t} = - 4 v^{3}$
So, the correct choice is (d).

Suggest Corrections

Similar questions

Q. The distance

x

covered by a body moving in a straight line in

t

is given by the relation

2 x^{2} + 3 x = t .

v

is the velocity of the body at a certain of time, its acceleration will be :

Q. The velocity V of a body moving along a straight line varies with time

t

v = 2 t^{2} e^{- t}

, where v is in m/s and

t

in second . The acceleration of body is zero at

t

Figure shows the acceleration – time ( $a - t$ ) graph of a body moving in a straight line. Which graph in figure shows the velocity – time ( $v - t$ ) graph of the motion of the body? (Assume that $x = 0$ and $v = 0$ at $t = 0$ )

Q. Assertion: A body is moving along a straight line such that its velocity varies with time as shown in figure. Magnitude of displacement of the body from

t = 0

t = 12

s is the same as the distance travelled by it in the given time duration.

Reason: For unidirectional motion of a body, |displacement|=distance.

Q. The velocity of a body moving along a straight varies with time as

V = t^{2} - 4 t m / s

, find the average acceleration of the body during an interval

t = 1 s e c

t = 3 s e c

Join BYJU'S Learning Program

The displacement x of a body moving in a straight line is related to time t as 2x2+3x=t. If v is the velocity of the body at a certain instant of time, its acceleration will be

The correct option is D −4v3 Differentiating 2x2+3x=t with respect to t, 4xdxdt+3dxdt=1 Now, dxdt=v ∴4xv+3v=1 ⇒4x+3=1v Diffrentiating with respect to time t, 4dxdt=−1v2dvdt ⇒4v=−1v2dvdt ⇒a=dvdt=−4v3 So, the correct choice is (d).

The displacement x of a body moving in a straight line is related to time t as $2 x^{2} + 3 x = t$ . If v is the velocity of the body at a certain instant of time, its acceleration will be