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

Question

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to: (A)  t1/2  (B)  t3/2  (C)  t2  (D)  t3/4

Tardigrade Admin · Accepted Answer

As power =F× upsilon =m(dv/dt)× v= constant k upsilon d upsilon =(k/m)dt ?(i) Now integrating equation (i), we get upsilon 2=(2kt/m) or upsilon =(dx/dt)=√(2k/m)t1/2 hence x propto t3/2

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

$t^{1/2}$

$t^{3/2}$

$t^{2}$

$t^{3/4}$

As power $= F \times υ = m \frac{d v}{d t} \times v =$ constant k $υ d υ = \frac{k}{m} d t$ ?(i) Now integrating equation (i), we get $υ^{2} = \frac{2 k t}{m}$ or $υ = \frac{d x}{d t} = \frac{2 k}{m} t^{1/2}$ hence $x \propto t^{3/2}$

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

t1/2

t3/2

t2

t3/4

As power =F×υ=mdtdv​×v= constant k υdυ=mk​dt ?(i) Now integrating equation (i), we get υ2=m2kt​ or υ=dtdx​=m2k​​t1/2 hence x∝t3/2

$t^{1/2}$

$t^{3/2}$

$t^{2}$

$t^{3/4}$

As power $= F \times υ = m \frac{d v}{d t} \times v =$ constant k $υ d υ = \frac{k}{m} d t$ ?(i) Now integrating equation (i), we get $υ^{2} = \frac{2 k t}{m}$ or $υ = \frac{d x}{d t} = \frac{2 k}{m} t^{1/2}$ hence $x \propto t^{3/2}$