148709
When a position dependent force $F=7-2 x+$ $3 x^{2} N$ acts on a small body of mass $2 \mathrm{~kg}$ and displaces it from $x=0$ to $x=5 \mathrm{~m}$, calculate the work done (in joules):
1 70
2 270
3 35
4 135
Explanation:
D Given, $F=7-2 x+3 x^{2}$ Work done is given by $\int_{0}^{\mathrm{W}} \mathrm{dw}=\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{~F} \cdot \mathrm{dx}$ $\mathrm{W}=\int_{0}^{5}\left(7-2 \mathrm{x}+3 \mathrm{x}^{2}\right) \mathrm{dx}$ $=\left[7 \mathrm{x}-\frac{2 \mathrm{x}^{2}}{2}+\frac{3 \mathrm{x}^{3}}{3}\right]_{0}^{5}$ $=\left[7 \mathrm{x}-\mathrm{x}^{2}+\mathrm{x}^{3}\right]_{0}^{5}$ $=\left[7 \times 5-5^{2}+5^{3}-0\right]$ $=35-25+125$ $=135 \mathrm{~J}$
AP EAMCET-23.09.2020
Work, Energy and Power
148710
A body is initially at rest. It undergoes onedimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to :
1 $\mathrm{t}^{1 / 2}$
2 $\mathrm{t}$
3 $t^{3 / 2}$
4 $\mathrm{t}^{2}$
Explanation:
B Suppose $\mathrm{m}=$ mass of the body $a=$ acceleration produced in the body $\mathrm{v}=$ velocity of the body $\mathrm{P}=$ power delivered to the body in a time $\mathrm{t}$ $\mathrm{v} =\mathrm{u}+\text { at }$ $\mathrm{v} =\mathrm{at}$ $\mathrm{F} =\mathrm{ma}=0]$ Force, $\mathrm{F}=\mathrm{ma}$ Power delivered to the body - $\mathrm{P}=\mathrm{F} \times \mathrm{v}$ $\mathrm{P}=\mathrm{ma} \times \text { at }$ $\mathrm{P}=\mathrm{ma}^{2} \mathrm{t}$ So, $\mathrm{P} \propto \mathrm{t}$
Karnataka CET-2020
Work, Energy and Power
148711
A uniform chain of mass $M$ and length $L$ is lying on a smooth horizontal table, with half of its length hanging down. The work done is pulling the entire chain up the table is
1 $\frac{\mathrm{MgL}}{2}$
2 $\frac{\mathrm{MgL}}{4}$
3 $\frac{\mathrm{MgL}}{8}$
4 $\frac{\mathrm{MgL}}{16}$
Explanation:
C Given, mass $=\mathrm{M}$, Length $=\mathrm{L}$ mass of small part of chain $=\mathrm{dm}$ $\mathrm{dm}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{dx}$ work done to pulling $(\mathrm{dm})$ mass $\mathrm{dw}=\mathrm{dm} \text { g.x }$ $\mathrm{dw}=\left(\frac{\mathrm{M}}{\mathrm{L}} \cdot \mathrm{dx}\right) \mathrm{g} \cdot \mathrm{x}$ $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{gxdx}$ By integrating $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \int_{0}^{\mathrm{L} / 2} \mathrm{xdx}$ $\mathrm{W}=\left.\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{x}^{2}}{2}\right|_{0} ^{\mathrm{L} / 2}$ $\mathrm{~W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{L}^{2} / 4}{2}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$ II $^{\text {nd }}$ Method- Centre of mass at point $\mathrm{M}$ As we know that, $\text { Work done }(\mathrm{W})=\text { Change in energy }$ $\mathrm{W}=\Delta \mathrm{E}$ $\mathrm{W}=\mathrm{Mg} \Delta l$ $\Delta l=\frac{1}{2} \times \frac{\mathrm{L}}{2}=\frac{\mathrm{L}}{4}$ Now, as per given condition, $\mathrm{W}=\frac{\mathrm{M}}{2} \times \mathrm{g} \times \frac{\mathrm{L}}{4}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$
AP EAMCET (23.09.2020) Shift-I
Work, Energy and Power
148712
A ball is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to
1 $t^{1 / 2}$
2 $t^{3 / 4}$
3 $t^{3 / 2}$
4 $t^{2}$
Explanation:
C For moving of a body along the straight line, machine need to deliver a constant power- $\mathrm{P}=$ Force $(\mathrm{F}) \times$ Velocity $(\mathrm{v})$ $\because \mathrm{F}=\mathrm{ma}$ and $\therefore \mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{1}{\mathrm{t}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)$ Acceleration, $\mathrm{a}=\frac{\mathrm{s}}{\mathrm{t}^{2}}$, velocity, $\mathrm{v}=$ at $(\because$ Initial velocity, $\mathrm{u}=0)$ From, $\mathrm{eq}^{\mathrm{n}}$. (i), $\mathrm{P}=\mathrm{ma} \times \mathrm{v}$ $\mathrm{P}= \mathrm{ma} \times \mathrm{at}$ $\mathrm{P}= \mathrm{ma}^{2} \mathrm{t}$ $\mathrm{P}=\frac{\mathrm{ms}^{2}}{\mathrm{t}^{3}}$ $\mathrm{Pt}^{3}=\mathrm{ms}^{2}$ For constant power and mass distance will be- $\mathrm{s}^{2} \propto \mathrm{t}^{3}$ $\mathrm{~s} \propto \mathrm{t}^{3 / 2}$ Where, $\mathrm{s}=$ distance moved by body $\mathrm{t}=$ time
148709
When a position dependent force $F=7-2 x+$ $3 x^{2} N$ acts on a small body of mass $2 \mathrm{~kg}$ and displaces it from $x=0$ to $x=5 \mathrm{~m}$, calculate the work done (in joules):
1 70
2 270
3 35
4 135
Explanation:
D Given, $F=7-2 x+3 x^{2}$ Work done is given by $\int_{0}^{\mathrm{W}} \mathrm{dw}=\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{~F} \cdot \mathrm{dx}$ $\mathrm{W}=\int_{0}^{5}\left(7-2 \mathrm{x}+3 \mathrm{x}^{2}\right) \mathrm{dx}$ $=\left[7 \mathrm{x}-\frac{2 \mathrm{x}^{2}}{2}+\frac{3 \mathrm{x}^{3}}{3}\right]_{0}^{5}$ $=\left[7 \mathrm{x}-\mathrm{x}^{2}+\mathrm{x}^{3}\right]_{0}^{5}$ $=\left[7 \times 5-5^{2}+5^{3}-0\right]$ $=35-25+125$ $=135 \mathrm{~J}$
AP EAMCET-23.09.2020
Work, Energy and Power
148710
A body is initially at rest. It undergoes onedimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to :
1 $\mathrm{t}^{1 / 2}$
2 $\mathrm{t}$
3 $t^{3 / 2}$
4 $\mathrm{t}^{2}$
Explanation:
B Suppose $\mathrm{m}=$ mass of the body $a=$ acceleration produced in the body $\mathrm{v}=$ velocity of the body $\mathrm{P}=$ power delivered to the body in a time $\mathrm{t}$ $\mathrm{v} =\mathrm{u}+\text { at }$ $\mathrm{v} =\mathrm{at}$ $\mathrm{F} =\mathrm{ma}=0]$ Force, $\mathrm{F}=\mathrm{ma}$ Power delivered to the body - $\mathrm{P}=\mathrm{F} \times \mathrm{v}$ $\mathrm{P}=\mathrm{ma} \times \text { at }$ $\mathrm{P}=\mathrm{ma}^{2} \mathrm{t}$ So, $\mathrm{P} \propto \mathrm{t}$
Karnataka CET-2020
Work, Energy and Power
148711
A uniform chain of mass $M$ and length $L$ is lying on a smooth horizontal table, with half of its length hanging down. The work done is pulling the entire chain up the table is
1 $\frac{\mathrm{MgL}}{2}$
2 $\frac{\mathrm{MgL}}{4}$
3 $\frac{\mathrm{MgL}}{8}$
4 $\frac{\mathrm{MgL}}{16}$
Explanation:
C Given, mass $=\mathrm{M}$, Length $=\mathrm{L}$ mass of small part of chain $=\mathrm{dm}$ $\mathrm{dm}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{dx}$ work done to pulling $(\mathrm{dm})$ mass $\mathrm{dw}=\mathrm{dm} \text { g.x }$ $\mathrm{dw}=\left(\frac{\mathrm{M}}{\mathrm{L}} \cdot \mathrm{dx}\right) \mathrm{g} \cdot \mathrm{x}$ $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{gxdx}$ By integrating $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \int_{0}^{\mathrm{L} / 2} \mathrm{xdx}$ $\mathrm{W}=\left.\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{x}^{2}}{2}\right|_{0} ^{\mathrm{L} / 2}$ $\mathrm{~W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{L}^{2} / 4}{2}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$ II $^{\text {nd }}$ Method- Centre of mass at point $\mathrm{M}$ As we know that, $\text { Work done }(\mathrm{W})=\text { Change in energy }$ $\mathrm{W}=\Delta \mathrm{E}$ $\mathrm{W}=\mathrm{Mg} \Delta l$ $\Delta l=\frac{1}{2} \times \frac{\mathrm{L}}{2}=\frac{\mathrm{L}}{4}$ Now, as per given condition, $\mathrm{W}=\frac{\mathrm{M}}{2} \times \mathrm{g} \times \frac{\mathrm{L}}{4}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$
AP EAMCET (23.09.2020) Shift-I
Work, Energy and Power
148712
A ball is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to
1 $t^{1 / 2}$
2 $t^{3 / 4}$
3 $t^{3 / 2}$
4 $t^{2}$
Explanation:
C For moving of a body along the straight line, machine need to deliver a constant power- $\mathrm{P}=$ Force $(\mathrm{F}) \times$ Velocity $(\mathrm{v})$ $\because \mathrm{F}=\mathrm{ma}$ and $\therefore \mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{1}{\mathrm{t}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)$ Acceleration, $\mathrm{a}=\frac{\mathrm{s}}{\mathrm{t}^{2}}$, velocity, $\mathrm{v}=$ at $(\because$ Initial velocity, $\mathrm{u}=0)$ From, $\mathrm{eq}^{\mathrm{n}}$. (i), $\mathrm{P}=\mathrm{ma} \times \mathrm{v}$ $\mathrm{P}= \mathrm{ma} \times \mathrm{at}$ $\mathrm{P}= \mathrm{ma}^{2} \mathrm{t}$ $\mathrm{P}=\frac{\mathrm{ms}^{2}}{\mathrm{t}^{3}}$ $\mathrm{Pt}^{3}=\mathrm{ms}^{2}$ For constant power and mass distance will be- $\mathrm{s}^{2} \propto \mathrm{t}^{3}$ $\mathrm{~s} \propto \mathrm{t}^{3 / 2}$ Where, $\mathrm{s}=$ distance moved by body $\mathrm{t}=$ time
148709
When a position dependent force $F=7-2 x+$ $3 x^{2} N$ acts on a small body of mass $2 \mathrm{~kg}$ and displaces it from $x=0$ to $x=5 \mathrm{~m}$, calculate the work done (in joules):
1 70
2 270
3 35
4 135
Explanation:
D Given, $F=7-2 x+3 x^{2}$ Work done is given by $\int_{0}^{\mathrm{W}} \mathrm{dw}=\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{~F} \cdot \mathrm{dx}$ $\mathrm{W}=\int_{0}^{5}\left(7-2 \mathrm{x}+3 \mathrm{x}^{2}\right) \mathrm{dx}$ $=\left[7 \mathrm{x}-\frac{2 \mathrm{x}^{2}}{2}+\frac{3 \mathrm{x}^{3}}{3}\right]_{0}^{5}$ $=\left[7 \mathrm{x}-\mathrm{x}^{2}+\mathrm{x}^{3}\right]_{0}^{5}$ $=\left[7 \times 5-5^{2}+5^{3}-0\right]$ $=35-25+125$ $=135 \mathrm{~J}$
AP EAMCET-23.09.2020
Work, Energy and Power
148710
A body is initially at rest. It undergoes onedimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to :
1 $\mathrm{t}^{1 / 2}$
2 $\mathrm{t}$
3 $t^{3 / 2}$
4 $\mathrm{t}^{2}$
Explanation:
B Suppose $\mathrm{m}=$ mass of the body $a=$ acceleration produced in the body $\mathrm{v}=$ velocity of the body $\mathrm{P}=$ power delivered to the body in a time $\mathrm{t}$ $\mathrm{v} =\mathrm{u}+\text { at }$ $\mathrm{v} =\mathrm{at}$ $\mathrm{F} =\mathrm{ma}=0]$ Force, $\mathrm{F}=\mathrm{ma}$ Power delivered to the body - $\mathrm{P}=\mathrm{F} \times \mathrm{v}$ $\mathrm{P}=\mathrm{ma} \times \text { at }$ $\mathrm{P}=\mathrm{ma}^{2} \mathrm{t}$ So, $\mathrm{P} \propto \mathrm{t}$
Karnataka CET-2020
Work, Energy and Power
148711
A uniform chain of mass $M$ and length $L$ is lying on a smooth horizontal table, with half of its length hanging down. The work done is pulling the entire chain up the table is
1 $\frac{\mathrm{MgL}}{2}$
2 $\frac{\mathrm{MgL}}{4}$
3 $\frac{\mathrm{MgL}}{8}$
4 $\frac{\mathrm{MgL}}{16}$
Explanation:
C Given, mass $=\mathrm{M}$, Length $=\mathrm{L}$ mass of small part of chain $=\mathrm{dm}$ $\mathrm{dm}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{dx}$ work done to pulling $(\mathrm{dm})$ mass $\mathrm{dw}=\mathrm{dm} \text { g.x }$ $\mathrm{dw}=\left(\frac{\mathrm{M}}{\mathrm{L}} \cdot \mathrm{dx}\right) \mathrm{g} \cdot \mathrm{x}$ $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{gxdx}$ By integrating $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \int_{0}^{\mathrm{L} / 2} \mathrm{xdx}$ $\mathrm{W}=\left.\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{x}^{2}}{2}\right|_{0} ^{\mathrm{L} / 2}$ $\mathrm{~W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{L}^{2} / 4}{2}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$ II $^{\text {nd }}$ Method- Centre of mass at point $\mathrm{M}$ As we know that, $\text { Work done }(\mathrm{W})=\text { Change in energy }$ $\mathrm{W}=\Delta \mathrm{E}$ $\mathrm{W}=\mathrm{Mg} \Delta l$ $\Delta l=\frac{1}{2} \times \frac{\mathrm{L}}{2}=\frac{\mathrm{L}}{4}$ Now, as per given condition, $\mathrm{W}=\frac{\mathrm{M}}{2} \times \mathrm{g} \times \frac{\mathrm{L}}{4}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$
AP EAMCET (23.09.2020) Shift-I
Work, Energy and Power
148712
A ball is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to
1 $t^{1 / 2}$
2 $t^{3 / 4}$
3 $t^{3 / 2}$
4 $t^{2}$
Explanation:
C For moving of a body along the straight line, machine need to deliver a constant power- $\mathrm{P}=$ Force $(\mathrm{F}) \times$ Velocity $(\mathrm{v})$ $\because \mathrm{F}=\mathrm{ma}$ and $\therefore \mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{1}{\mathrm{t}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)$ Acceleration, $\mathrm{a}=\frac{\mathrm{s}}{\mathrm{t}^{2}}$, velocity, $\mathrm{v}=$ at $(\because$ Initial velocity, $\mathrm{u}=0)$ From, $\mathrm{eq}^{\mathrm{n}}$. (i), $\mathrm{P}=\mathrm{ma} \times \mathrm{v}$ $\mathrm{P}= \mathrm{ma} \times \mathrm{at}$ $\mathrm{P}= \mathrm{ma}^{2} \mathrm{t}$ $\mathrm{P}=\frac{\mathrm{ms}^{2}}{\mathrm{t}^{3}}$ $\mathrm{Pt}^{3}=\mathrm{ms}^{2}$ For constant power and mass distance will be- $\mathrm{s}^{2} \propto \mathrm{t}^{3}$ $\mathrm{~s} \propto \mathrm{t}^{3 / 2}$ Where, $\mathrm{s}=$ distance moved by body $\mathrm{t}=$ time
148709
When a position dependent force $F=7-2 x+$ $3 x^{2} N$ acts on a small body of mass $2 \mathrm{~kg}$ and displaces it from $x=0$ to $x=5 \mathrm{~m}$, calculate the work done (in joules):
1 70
2 270
3 35
4 135
Explanation:
D Given, $F=7-2 x+3 x^{2}$ Work done is given by $\int_{0}^{\mathrm{W}} \mathrm{dw}=\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{~F} \cdot \mathrm{dx}$ $\mathrm{W}=\int_{0}^{5}\left(7-2 \mathrm{x}+3 \mathrm{x}^{2}\right) \mathrm{dx}$ $=\left[7 \mathrm{x}-\frac{2 \mathrm{x}^{2}}{2}+\frac{3 \mathrm{x}^{3}}{3}\right]_{0}^{5}$ $=\left[7 \mathrm{x}-\mathrm{x}^{2}+\mathrm{x}^{3}\right]_{0}^{5}$ $=\left[7 \times 5-5^{2}+5^{3}-0\right]$ $=35-25+125$ $=135 \mathrm{~J}$
AP EAMCET-23.09.2020
Work, Energy and Power
148710
A body is initially at rest. It undergoes onedimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to :
1 $\mathrm{t}^{1 / 2}$
2 $\mathrm{t}$
3 $t^{3 / 2}$
4 $\mathrm{t}^{2}$
Explanation:
B Suppose $\mathrm{m}=$ mass of the body $a=$ acceleration produced in the body $\mathrm{v}=$ velocity of the body $\mathrm{P}=$ power delivered to the body in a time $\mathrm{t}$ $\mathrm{v} =\mathrm{u}+\text { at }$ $\mathrm{v} =\mathrm{at}$ $\mathrm{F} =\mathrm{ma}=0]$ Force, $\mathrm{F}=\mathrm{ma}$ Power delivered to the body - $\mathrm{P}=\mathrm{F} \times \mathrm{v}$ $\mathrm{P}=\mathrm{ma} \times \text { at }$ $\mathrm{P}=\mathrm{ma}^{2} \mathrm{t}$ So, $\mathrm{P} \propto \mathrm{t}$
Karnataka CET-2020
Work, Energy and Power
148711
A uniform chain of mass $M$ and length $L$ is lying on a smooth horizontal table, with half of its length hanging down. The work done is pulling the entire chain up the table is
1 $\frac{\mathrm{MgL}}{2}$
2 $\frac{\mathrm{MgL}}{4}$
3 $\frac{\mathrm{MgL}}{8}$
4 $\frac{\mathrm{MgL}}{16}$
Explanation:
C Given, mass $=\mathrm{M}$, Length $=\mathrm{L}$ mass of small part of chain $=\mathrm{dm}$ $\mathrm{dm}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{dx}$ work done to pulling $(\mathrm{dm})$ mass $\mathrm{dw}=\mathrm{dm} \text { g.x }$ $\mathrm{dw}=\left(\frac{\mathrm{M}}{\mathrm{L}} \cdot \mathrm{dx}\right) \mathrm{g} \cdot \mathrm{x}$ $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{gxdx}$ By integrating $\mathrm{dw}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \int_{0}^{\mathrm{L} / 2} \mathrm{xdx}$ $\mathrm{W}=\left.\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{x}^{2}}{2}\right|_{0} ^{\mathrm{L} / 2}$ $\mathrm{~W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{g} \frac{\mathrm{L}^{2} / 4}{2}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$ II $^{\text {nd }}$ Method- Centre of mass at point $\mathrm{M}$ As we know that, $\text { Work done }(\mathrm{W})=\text { Change in energy }$ $\mathrm{W}=\Delta \mathrm{E}$ $\mathrm{W}=\mathrm{Mg} \Delta l$ $\Delta l=\frac{1}{2} \times \frac{\mathrm{L}}{2}=\frac{\mathrm{L}}{4}$ Now, as per given condition, $\mathrm{W}=\frac{\mathrm{M}}{2} \times \mathrm{g} \times \frac{\mathrm{L}}{4}$ $\mathrm{~W}=\frac{\mathrm{MgL}}{8}$
AP EAMCET (23.09.2020) Shift-I
Work, Energy and Power
148712
A ball is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to
1 $t^{1 / 2}$
2 $t^{3 / 4}$
3 $t^{3 / 2}$
4 $t^{2}$
Explanation:
C For moving of a body along the straight line, machine need to deliver a constant power- $\mathrm{P}=$ Force $(\mathrm{F}) \times$ Velocity $(\mathrm{v})$ $\because \mathrm{F}=\mathrm{ma}$ and $\therefore \mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}}=\frac{1}{\mathrm{t}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)$ Acceleration, $\mathrm{a}=\frac{\mathrm{s}}{\mathrm{t}^{2}}$, velocity, $\mathrm{v}=$ at $(\because$ Initial velocity, $\mathrm{u}=0)$ From, $\mathrm{eq}^{\mathrm{n}}$. (i), $\mathrm{P}=\mathrm{ma} \times \mathrm{v}$ $\mathrm{P}= \mathrm{ma} \times \mathrm{at}$ $\mathrm{P}= \mathrm{ma}^{2} \mathrm{t}$ $\mathrm{P}=\frac{\mathrm{ms}^{2}}{\mathrm{t}^{3}}$ $\mathrm{Pt}^{3}=\mathrm{ms}^{2}$ For constant power and mass distance will be- $\mathrm{s}^{2} \propto \mathrm{t}^{3}$ $\mathrm{~s} \propto \mathrm{t}^{3 / 2}$ Where, $\mathrm{s}=$ distance moved by body $\mathrm{t}=$ time
