A point moves such that its displacement as a function of time is given by x^3 = t^3 + 1 . Its accel

English

Gujarati

Hindi

2.Motion in Straight Line

medium

A point moves such that its displacement as a function of time is given by $x^3$ = $t^3 + 1$. Its acceleration as a function of time $t$ will be

A$\frac{2}{x^5}$

B$\frac{2t}{x^5}$

C$\frac{2t}{x^4}$

D$\frac{2t^2}{x^5}$

Solution

$x^{3}=t^{3}+1 \Rightarrow 3 x^{2} \frac{d x}{d t}=3 t^{2}$
$\mathrm{x}^{2} \mathrm{v}=\mathrm{t}^{2} \Rightarrow 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}} \mathrm{v}+\mathrm{x}^{2} \mathrm{a}=2 \mathrm{t}$
$2 \mathrm{x} \frac{\mathrm{t}^{4}}{\mathrm{x}^{4}}+\mathrm{x}^{2} \mathrm{a}=2 \mathrm{t} \Rightarrow \mathrm{x}^{2} \mathrm{a}=2 \mathrm{t}-\frac{2 \mathrm{t}^{4}}{\mathrm{x}^{3}}$
$a=\frac{2 t\left[x^{3}-t^{3}\right]}{x^{5}} \Rightarrow a=\frac{2 t}{x^{5}}$

Standard 11

Physics

The acceleration-time graph of a body is shown below The most probable velocity-time graph of the body is

hard

Match the following columns.

Colum $I$ Colum $II$

$(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration

$(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration

$(C)$ $\frac{dr}{dt}$ $(r)$ Velocity

$(D)$ $\left|\frac{d r }{d t}\right|$ $(s)$ Magnitude of velocity

easy

The acceleration (a)-time $(t)$ graph for a particle moving along a straight starting from rest is shown in figure. Which of the following graph is the best representation of variation of its velocity $(v)$ with time $(t)$ ?

hard

The displacement of a particle moving in a straight line depends on time as $x=\alpha t^3+\beta t^2+\gamma t+\delta$.The ratio of initial acceleration to its initial velocity depends

medium

$Assertion$ : A body can have acceleration even if its velocity is zero at a given instant of time.

$Reason$ : A body is numerically at rest when it reverses its direction.

easy

(AIIMS-1998)

Colum $I$	Colum $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d\|v\|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left\|\frac{d r }{d t}\right\|$	$(s)$ Magnitude of velocity

A point moves such that its displacement as a function of time is given by $x^3$ = $t^3 + 1$. Its acceleration as a function of time $t$ will be

Solution

Similar Questions

The acceleration-time graph of a body is shown below The most probable velocity-time graph of the body is

Match the following columns. Colum $I$ Colum $II$ $(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration $(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration $(C)$ $\frac{dr}{dt}$ $(r)$ Velocity $(D)$ $\left|\frac{d r }{d t}\right|$ $(s)$ Magnitude of velocity

The acceleration (a)-time $(t)$ graph for a particle moving along a straight starting from rest is shown in figure. Which of the following graph is the best representation of variation of its velocity $(v)$ with time $(t)$ ?

The displacement of a particle moving in a straight line depends on time as $x=\alpha t^3+\beta t^2+\gamma t+\delta$.The ratio of initial acceleration to its initial velocity depends

$Assertion$ : A body can have acceleration even if its velocity is zero at a given instant of time. $Reason$ : A body is numerically at rest when it reverses its direction.

Start a Free Trial Now

Match the following columns.

Colum $I$ Colum $II$

$(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration

$(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration

$(C)$ $\frac{dr}{dt}$ $(r)$ Velocity

$(D)$ $\left|\frac{d r }{d t}\right|$ $(s)$ Magnitude of velocity

$Assertion$ : A body can have acceleration even if its velocity is zero at a given instant of time.

$Reason$ : A body is numerically at rest when it reverses its direction.