Question

A particle moves along a curve so that its coordinates at time $t$ are $x=t,y=\frac{1}{2}{t}^2,z=\frac{1}{3}{t}^3$ acceleration at $t=1$ is

• A. $j+2k$
• B. $j+k$
• C. $2j+k$
• D. none of these

Answer 1

The correct option is A $j+2k$

Let $x,y,z$ be the magnitudes of the displacements along the three axes $X,Y,Z$ respectively.

Let $V_x,V_y,V_z$ be the respective velocities and $a_x,a_y,a_z$ be the respective accelerations along the axes $X,Y,Z$ respectively.

$x=t,y=\frac{1}{2}{t}^2,z=\frac{1}{3}{t}^3$

$V_x=\frac{dx}{dt}=1,V_y=\frac{dy}{dt}=t,V_z=\frac{dz}{dt}=t^2$

$a_x=\frac{d{V}_x}{dt}=0,a_y=\frac{d{V}_y}{dt}=1,a_z=\frac{d{V}_z}{dt}=2t$

$\therefore$ acceleration at $t=1$ is $j+2k$