Question

The position vector of a particle changes with time according to the relation $\vec{r} (t) = 15t^2 \hat{i} + ( 4 -20 t^2) \hat{j}$. What is the magnitude of the acceleration at $t = 1$ ?

• A. 40
• B. 100
• C. 25
• D. 50

Answer 1

Answer: (D)

$\vec{r} = 15t^{2} \hat{i} +\left(4-20 t^{2}\right)\hat{j} $
$ \vec{v} = \frac{d\vec{r}}{dt} =30 t \hat{i} + \left(-40t\right)\hat{j} $
$ \vec{a} = \frac{d \vec{v}}{dt} =30 \hat{i} - 40 \hat{j}$
$ \left|\vec{a}\right| = 50 m/s^{2} $