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Q. An equation $ x=t^{3}-2f $ denotes relationship between displacement and time. At $t = 4 s$, acceleration is given by:

J & K CETJ & K CET 2001

Solution:

Rate of change of velocity gives acceleration.
Given, $x=t^{3}-2 t$ Differentiating with respect to $t$,
we get $v=\frac{d x}{d t}=3 t^{2}-2$
For acceleration $v=\frac{d x}{d t}=\frac{d^{2} x}{d t^{2}}=6 t$
Given, $t=4 \,s $
$\therefore a=6 \times 4=24 \,m / s ^{2}$