Question

the acceleration of a particle as a function of time t is given as a=k.t^5/2. if initial speed of the particle at t=0 is u then its velocity v as a function of time t is given as

Answer 1

Explanation:

explanation attached

Answer 2

The velocity 'v' of the particle as a function of time 't' can be represented as: $\boldsymbol{v=\frac{2k}{7}(t^{\frac{7}{2} })}$.

Given:

Acceleration-time function: $\boldsymbol{a=kt^{\frac{5}{2} } }$

The initial speed of the particle at t = 0 is zero.

To Find:

The velocity (v) as a function of time (t).

Solution:

→ The acceleration of a particle can be defined as the rate of change of its velocity.

We can calculate the acceleration of a body by dividing the net change in velocity (Δv) by the total time taken (Δt).          

                                               $\boldsymbol{\therefore a=\frac{\triangle v}{\triangle t} }$                    

→ The differential form: The differentiation of the velocity (v) of the body with respect to time (t) gives the acceleration (a) of the body.

                                               $\boldsymbol{\therefore a=\frac{d v}{d t} }$

→ In the given question:

→ The initial velocity of the particle at t = 0 is zero.

Assume: The velocity of the particle after time 't' be 'v'.

The acceleration-time function: $\boldsymbol{a=kt^{\frac{5}{2} } }$

                                                $\therefore \frac{dv}{dt} =kt^{\frac{2}{5} } \\\\\therefore dv=kt^{\frac{2}{5} }dt$

→ Integrating both sides with proper limit:

                                             $\therefore \int\limits^v_0 {} \, dv =\int\limits^t_0 {kt^{\frac{5}{2} } } \, dt \\\\\therefore \int\limits^v_0 {} \, dv =k\int\limits^t_0 {t^{\frac{5}{2} } } \, dt \\\\\therefore (v-0)=\frac{k}{(\frac{5}{2} +1)} t^{\frac{7}{2} } \\\\\therefore v=\frac{2k}{7}(t^{\frac{7}{2} })$

Therefore the velocity 'v' of the particle as a function of time 't' can be represented as: $\boldsymbol{v=\frac{2k}{7}(t^{\frac{7}{2} })}$.

#SPJ2