Question

The acceleration of a particle as a function of time is given by a=1.5t-0.15t^2.the particle starts motion from rest at t=0 then maximum velocity in forward direction

Answer 1

Answer:

1.h=ut+1/2at^2

2.v=u+at

Answer 2

The maximum velocity in forwarding direction is 22.22 m/sec

Equations of motion in physics are equations that explain how a physical system behaves in terms of how its motion changes over time. The behavior of a physical system is described in more detail by the equations of motion as a collection of mathematical functions expressed in terms of dynamic variables. These variables typically consist of space, time, and sometimes elements of momentum.

The three equations of motion are as follows-

$v=u+at$

$v^{2}=u^{2}+2as$

$s=ut+\frac{1}{2}at^{2}$

where s is the displacement, u is the beginning velocity, v is the final velocity, an is the acceleration, and t is the length of the motion. SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time, and these equations are known as SUVAT equations (T).

Given,

at t=0,

$a=1.5t-0.15t^2$

the particle starts from rest that is,

initial velocity(u) = 0

By applying the first equation of motion the velocity at time t is given as,

$v=u+(1.5t-0.15t^{2})t$

⇒ $v=0+1.5t^{2}-0.15t^{3}$

⇒$v=1.5t^{2}-0.15t^{3}$

So, the maximum velocity is given by $\frac{dv}{dt}=0$

⇒ $\frac{dv}{dt} = 1.5\times 2t -0.15\times 3 t^{2}$

⇒ $\frac{dv}{dt} = 3t-0.45t^{2}$

⇒ $3t-0.45t^{2}=0$

⇒ $t(3-0.45t)=0$

⇒ $t=0$

and $3-0.45t=0$

⇒ $0.45t=3$

⇒ $t=\frac{3}{0.45}$

$t=6.67 sec$

Therefore, the maximum velocity is-

$v=1.5(6.67)^{2}-0.15(6.67)^{3}$

$v=66.73-44.51$

$v=22.22 msec^{-1}$

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