A point moves in a straight line so that its | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$x^{2}=1+t^{2}$

$	ext { or } \quad x=\left(1+t^{2}ight)^{1 / 2}$

${\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}\left(1+\mathrm{t}^{2}ight)^{

Vedclass · Accepted Answer

$\frac{1}{x} - \frac{{{t^2}}}{{{x^3}}}$

Vedclass · Answer

$x^{2}=1+t^{2}$

$	ext { or } \quad x=\left(1+t^{2}ight)^{1 / 2}$

${\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}\left(1+\mathrm{t}^{2}ight)^{-1 / 2} \cdot 2 \mathrm{t}=\mathrm{t}\left(1+\mathrm{t}^{2}ight)^{-1 / 2}}$

${\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\mathrm{t}\left(-\frac{1}{2}ight)\left(1+\mathrm{t}^{2}ight)^{-3 / 2} \cdot 2 \mathrm{t}+\left(1+\mathrm{t}^{2}ight)^{-1 / 2}}$

${=\frac{1}{\mathrm{x}}-\frac{\mathrm{t}^{2}}{\mathrm{x}^{3}}}$

A point moves in a straight line so that its displacement $x\,m$ at time $t\,sec$ is given by $x^2 = 1 + t^2$. Its acceleration in $m/sec^2$ at a time $t\,sec$ is

Similar Questions

$Assertion$ : Retardation is directly opposite to the velocity.
$Reason$ : Retardation is equal to the time rate of decrease of speed.

From the $v-t$ graph, the

For the acceleration-time $(a-t)$ graph shown in figure, the change in velocity of particle from $t=0$ to $t=6 \,s$ is ........ $m / s$

A car moving along a straight highway with speed of $126 \;\mathrm{km} h^{-1}$ is brought to a stop within a distance of $200\; \mathrm{m}$. how long(in $seconds$) does it take for the car to stop?

The displacement of a particle is given by $y = a + bt + c{t^2} - d{t^4}$. The initial velocity and acceleration are respectively