A point moves such that its displacement as function of time is given by x2=t2+1. Its acceleration at time t is

Question

A point moves such that its displacement as function of time is given by x2=t2+1. Its acceleration at time t is (A)  (1/x4)  (B)  -(t/x2)  (C)  (1/x)-(t2/x3)  (D)  (1/x)-(t/x2)

Tardigrade Admin · Accepted Answer

x2=t2+t Differentiate wrt x 2× (dx/dt)=2t v=(dx/dt)=(t/x) a=(dv/dt) =(x-t( (dx/dt) )/x2) =(x-( (t2/x) )/x2) a=(1/x)-(t2/x3)

Q. A point moves such that its displacement as function of time is given by $x^{2} = t^{2} + 1.$ Its acceleration at time $t$ is

$\frac{1}{x ^{4}}$

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

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

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

$x^{2} = t^{2} + t$ Differentiate wrt x $2 \times \frac{d x}{d t} = 2 t$ $v = \frac{d x}{d t} = \frac{t}{x}$ $a = \frac{d v}{d t}$ $= \frac{x - t ( \frac{d x}{d t} )}{x ^{2}}$ $= \frac{x - ( \frac{t ^{2}}{x} )}{x ^{2}}$ $a = \frac{1}{x} - \frac{t ^{2}}{x ^{3}}$

Q. A point moves such that its displacement as function of time is given by x2=t2+1. Its acceleration at time t is

x41​

−x2t​

x1​−x3t2​

x1​−x2t​