Question

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to

Answer 1

Answer:

Explanation:

Let's consider a body is moved along a straight line by a machine delivering a constant power P. The distance moved by the body is S.  

Power, P=F.v.. . . . . . (1)

Force, F=ma . . . . . . . .(2)

where, v=  

t

S

​  

 

acceleration, a=  

t  

2

 

S

​  

 

m= mass

from  equation (1) and equation (2), we get

P=  

t  

2

 

mS

​  

×  

t

S

​  

 

S  

2

=  

m

Pt  

3

 

​  

 

From the above equation, we get

S  

2

∝t  

3

 

S∝t  

3/2

 

Answer 2

$\boxed{ \boxed {\sf \red{Key \: Idea}}}$

In case of velocity changing with distance, we apply integration method to determine the relationship between distance and time, if force is given

$\boxed{ \boxed {\sf \red{Let \: us \: come \: to \: question}}}$

$\sf{As, \: Power, P = constant}$

$\implies \sf \: Fv = P \: \: \: \: \: \: \: \: (\because P = force \times velocity)$

$\implies \sf \: Ma \times v = P \: \: \: \: \: \: \: \: \: \: (\because F = Ma)$

$\implies \sf\: va = \frac{P}{M}$

$\implies \sf \: v \times \Bigg[ \frac{v \: dv}{ds}\Bigg ] = \frac{P}{M} \: \: \: \: \: \Bigg( \because \: a = \frac{v \: dv}{ds} \Bigg)$

$\implies \sf \int \limits ^{v} _0 \: {v}^{2} \: dv = \int \limits ^{s} _0 \frac{P}{M} \: ds$

(assuming at t = 0 it starts from rest, i.e., from s = 0)

$\implies \sf \frac{ {v}^{3} }{3} = \frac{P}{M} \: s$

$\implies \sf \: v \: = \bigg( \frac{3P}{M} \bigg)^{1/3}. {s}^{1/3}$

$\implies \sf \frac{ds}{dt} = K \: {s}^{1/3} \: \: \: \: \: \: \: \: \: \: \: \Bigg[ \because \: K = \bigg( \frac{3P}{M} \bigg)^{1/3} \Bigg]$

$\implies \sf \int \limits ^{s} _0 \frac{ds}{ {s}^{1/3 } } = \int \limits ^{t} _0 \: K \: dt$

$\implies \sf \frac{ {s}^{2/3 } }{2/3 } = Kt$

$\implies \sf \: {s}^{2/3 } = \frac{2}{3} Kt$

$\implies \sf s = \bigg( \frac{2}{3} K \bigg)^{3/2} \: \times \: {t}^{3/2}$

$\implies \boxed{ \boxed{\: s \propto \: {t}^{3/2} }}$