Question

1.24. A radius vector of a point A relative to the origin varies with
time t as r = ati — bt2 j, where a and b are positive constants, and i
and j are the unit vectors of the x and y axes. Find:
(a) the equation of the point's trajectory y (x); plot this function;
(b) the time dependence of the velocity v and acceleration w vectors,
as well as of the moduli of these quantities;
(c) the time dependence of the angle a between the vectors w and v;
(d) the mean velocity vector averaged over the first t seconds of
motion, and the modulus of this vector.

Answer 1

$\large\underline{\bold{Solution-}}$

$\underline{\bf \bold{ \: Answer \: (a)}}$

It is given that

$\rm :\longmapsto\:\vec{r} = a t \: \hat{i} - b {t}^{2} \: \hat{j}$

So,

• On comparing with,

$\rm :\longmapsto\:\vec{r} = x \: \hat{i} + y \: \hat{j}$

• we get

$\rm :\implies\:x = at \: \: and \: \: y = - b {t}^{2}$

• Now, eliminate 't' from these two conditions,

$\rm :\longmapsto\: \: as \: x = at \: \bf :\implies\:t = \dfrac{x}{a}$

So,

$\rm :\implies\:y = - b \times \dfrac{ {x}^{2} }{ {a}^{2} }$

$\bf :\implies\:y = - \dfrac{b {x}^{2} }{ {a}^{2} }$

• which is the equation of Parabola whose vertex is at the origin and shape downward.

• . Please see the attachment for graph

$\underline{\bf \bold{ \: Answer \: (b)}}$

We know,

$\rm :\longmapsto\:\vec{v} \: = \: \dfrac{d}{dt} \vec{r}$

and

$\rm :\longmapsto\:\vec{w} = \dfrac{d}{dt} \vec{v}$

As,

$\rm :\longmapsto\:\vec{r} = a t \: \hat{i} - b {t}^{2} \: \hat{j}$

• On differentiating with respect to 't', we get

$\rm :\longmapsto\:\dfrac{d}{dr} \vec{r} = \dfrac{d}{dt} (a t \: \hat{i} - b {t}^{2} \: \hat{j})$

$\bf :\longmapsto\:\vec{v} = a \: \hat{i} - 2b {t} \: \hat{j}$

• On differentiating with respect to 't', we get

$\rm :\longmapsto\:\dfrac{d}{dt} \vec{v} = \dfrac{d}{dt} (a\: \hat{i} - 2b {t} \: \hat{j})$

$\bf :\longmapsto\:\vec{w} = - 2b \: \hat{j}$

Now,

$\rm :\longmapsto\: |\vec{v}| = \sqrt{ {a}^{2} + {( - 2bt)}^{2} }$

$\bf\implies \: |\vec{v}| = \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} }$

And

$\rm :\longmapsto\: |\vec{w}| = \sqrt{ {( - 2b)}^{2} }$

$\bf :\longmapsto\: |\vec{w}| = 2b$

$\underline{\bf \bold{ \: Answer \: (c)}}$

We know,

To distinguish the angle 'a' with numeric constant 'a',

• Let angle between two vectors be 'p'.

• Angle between two vector is given by

$\rm :\longmapsto\:cosp \: = \dfrac{\vec{v} \: . \: \vec{w}}{ |\vec{v}| \: |\vec{w}| }$

$\rm :\longmapsto\:cosp \: = \dfrac{(a\: \hat{i} - 2bt \: \hat{j}) \: . \: ( - 2b \: \hat{j})}{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } \times 2b }$

$\rm :\longmapsto\:cosp = \dfrac{ {4tb}^{2} }{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } \times 2b }$

$\bf\implies \:cosp = \dfrac{2bt}{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } }$

As,

We know

$\rm :\longmapsto\: {tan}^{2} p = {sec}^{2} p - 1$

$\rm :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2} + 4 {b}^{2} {t}^{2} }{4 {b}^{2} {t}^{2} } - 1$

$\rm :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2} + 4 {b}^{2} {t}^{2} - 4 {b}^{2} {t}^{2}}{4 {b}^{2} {t}^{2} }$

$\rm :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2}}{4 {b}^{2} {t}^{2} }$

$\rm :\longmapsto\:tanp \: = \: \dfrac{a}{2bt}$

$\bf\implies \:p \: = \: {tan}^{ - 1} \bigg(\dfrac{a}{2bt} \bigg)$

$\underline{\bf \bold{ \: Answer \: (d)}}$

The mean velocity vector is given by

$\rm :\longmapsto\:\vec{ \overline{v}} = \dfrac{\int_0^t \: \vec{v} \: dt}{ \int \: dt}$

$\rm :\longmapsto\:\vec{ \overline{v}} = \dfrac{\int_0^t \: (a\: \hat{i} - 2bt \: \hat{j}) \: dt}{ t}$

$\rm :\longmapsto\:\vec{ \overline{v}} = \dfrac{at\: \hat{i} - 2b \: \dfrac{ {t}^{2} }{2} \: \hat{j} }{t}$

$\bf :\longmapsto\:\vec{ \overline{v}} = a\: \hat{i} - b {t} \: \hat{j}$

Hence,

$\rm :\longmapsto\: |\vec{ \overline{v}} | = \sqrt{ {a}^{2} + {( - bt)}^{2} }$

$\bf\implies \: |\vec{ \overline{v}}| = \sqrt{ {a}^{2} + {b}^{2} {t}^{2} }$

Answer 2

$\large\underline{\bold{\red{Solution\downarrow}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\underline{\bf \bold{ \:\pink{ Answer \: (a)\downarrow}}}$

It is given that

$\rm :\longmapsto\:\vec{r} = a t \: \hat{i} - b {t}^{2} \: \hat{j}$

So,

• On comparing with,

$\rm :\longmapsto\:\vec{r} = x \: \hat{i} + y \: \hat{j}$

• we get

$\rm :\implies\:x = at \: \: and \: \: y = - b {t}^{2}$

• Now, eliminate 't' from these two conditions,

$\rm :\longmapsto\: \: as \: x = at \: \bf :\implies\:t = \dfrac{x}{a}$

So,

$\rm :\implies\:y = - b \times \dfrac{ {x}^{2} }{ {a}^{2} }$

$\bf :\implies\:y = - \dfrac{b {x}^{2} }{ {a}^{2} }$

• which is the equation of Parabola whose vertex is at the origin and shape downward.
• Please see the attachment for graph.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\underline{\bf \bold{ \:\green{ Answer \: (b)\downarrow}}}$

We know,

$\rm :\longmapsto\:\vec{v} \: = \: \dfrac{d}{dt} \vec{r}$

and

$\rm :\longmapsto\:\vec{w} = \dfrac{d}{dt} \vec{v}$

As,

$\rm :\longmapsto\:\vec{r} = a t \: \hat{i} - b {t}^{2} \: \hat{j}$

• On differentiating with respect to 't', we get

$\rm :\longmapsto\:\dfrac{d}{dr} \vec{r} = \dfrac{d}{dt} (a t \: \hat{i} - b {t}^{2} \: \hat{j})$

$\bf :\longmapsto\:\vec{v} = a \: \hat{i} - 2b {t} \: \hat{j}$

• On differentiating with respect to 't', we get

$\rm :\longmapsto\:\dfrac{d}{dt} \vec{v} = \dfrac{d}{dt} (a\: \hat{i} - 2b {t} \: \hat{j})$

$\bf :\longmapsto\:\vec{w} = - 2b \: \hat{j}$

Now,

$\rm :\longmapsto\: |\vec{v}| = \sqrt{ {a}^{2} + {( - 2bt)}^{2} }$

$\bf\implies \: |\vec{v}| = \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} }$

And

$\rm :\longmapsto\: |\vec{w}| = \sqrt{ {( - 2b)}^{2} }$

$\bf :\longmapsto\: |\vec{w}| = 2b$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\underline{\bf \bold{ \:\purple{ Answer \: (c)\downarrow}}}$

We know,

To distinguish the angle 'a' with numeric constant 'a',

• Let angle between two vectors be 'p'.
• Angle between two vector is given by

$\rm :\longmapsto\:cosp \: = \dfrac{\vec{v} \: . \: \vec{w}}{ |\vec{v}| \: |\vec{w}| }$

$\rm :\longmapsto\:cosp \: = \dfrac{(a\: \hat{i} - 2bt \: \hat{j}) \: . \: ( - 2b \: \hat{j})}{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } \times 2b }$

$\rm :\longmapsto\:cosp = \dfrac{ {4tb}^{2} }{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } \times 2b }$

$\bf\implies \:cosp = \dfrac{2bt}{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } }$

As,

We know

$\rm :\longmapsto\: {tan}^{2} p = {sec}^{2} p - 1$

$\rm :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2} + 4 {b}^{2} {t}^{2} }{4 {b}^{2} {t}^{2} } - 1$

$\rm :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2} + 4 {b}^{2} {t}^{2} - 4 {b}^{2} {t}^{2}}{4 {b}^{2} {t}^{2} }$

$\rm :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2}}{4 {b}^{2} {t}^{2} }$

$\rm :\longmapsto\:tanp \: = \: \dfrac{a}{2bt}$

$\bf\implies \:p \: = \: {tan}^{ - 1} \bigg(\dfrac{a}{2bt} \bigg)$

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$\underline{\bf \bold{ \:\orange{ Answer \: (d)\downarrow}}}$

The mean velocity vector is given by

$\rm :\longmapsto\:\vec{ \overline{v}} = \dfrac{\int_0^t \: \vec{v} \: dt}{ \int \: dt}$

$\rm :\longmapsto\:\vec{ \overline{v}} = \dfrac{\int_0^t \: (a\: \hat{i} - 2bt \: \hat{j}) \: dt}{ t}$

$\rm :\longmapsto\:\vec{ \overline{v}} = \dfrac{at\: \hat{i} - 2b \: \dfrac{ {t}^{2} }{2} \: \hat{j} }{t}$

$\bf :\longmapsto\:\vec{ \overline{v}} = a\: \hat{i} - b {t} \: \hat{j}$

Hence,

$\rm :\longmapsto\: |\vec{ \overline{v}} | = \sqrt{ {a}^{2} + {( - bt)}^{2} }$

$\bf\implies \: |\vec{ \overline{v}}| = \sqrt{ {a}^{2} + {b}^{2} {t}^{2} }$

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