If position vector of a particle is vec r = - cos ,that{i} + sin ,that{j}

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3-2.Motion in Plane

medium

If the position vector of a particle is

$\vec r = - \cos \,t\hat i + \sin \,t\hat j - 18\,t\hat k$

then what is the magnitude of its acceleration ?

$0$

$1$

$sin^2\,t$

$cos\,t$

Solution

$\overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=\sin {\mathrm{t} \hat{\mathrm{i}}}+\cos \hat{\mathrm{t} {\mathrm{j}}}-18 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{V}}}{\mathrm{dt}}=\cos t \hat{\mathrm{i}}-\sin t \hat{\mathrm{j}}$

$|\overrightarrow{\mathrm{a}}|=\sqrt{\cos ^{2} t+(-\sin t)^{2}}=1$

Standard 11

Physics

A particle $(A)$ is dropped from a height and another particle $(B)$ is thrown in horizontal direction with speed of $5\; m/sec$ from the same height. The correct statement is

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(AIPMT-2002)

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$(a)$ $\left. v _{\text {average }}=(1 / 2) \text { (v }\left(t_{1}\right)+ v \left(t_{2}\right)\right)$

$(b)$ $v _{\text {average }}=\left[ r \left(t_{2}\right)- r \left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$

$(c)$ $v (t)= v (0)+ a t$

$(d)$ $r (t)= r (0)+ v (0) t+(1 / 2)$ a $t^{2}$

$(e)$ $a _{\text {merage }}=\left[ v \left(t_{2}\right)- v \left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$

(The 'average' stands for average of the quantity over the time interval $t_{1}$ to $t_{2}$ )

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If the position vector of a particle is $\vec r = - \cos \,t\hat i + \sin \,t\hat j - 18\,t\hat k$ then what is the magnitude of its acceleration ?

Solution

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If the position vector of a particle is

$\vec r = - \cos \,t\hat i + \sin \,t\hat j - 18\,t\hat k$

then what is the magnitude of its acceleration ?