Question

A particle is moving with a velocity of v =(3+6t+9t^2)cm/s. Find out
1. The acceleration of the particle at t=3s
2. The displacement of the particle in the interval t=5s to t=8s.

Answer 1

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Answer 2

The acceleration and displacement is $60cm/s^2$ and $1287m$.

Given-Velocity of a particle is $3+6t+9t^2cm/s$

Find the acceleration of the particle. The displacement of the particle

In mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

Velocity is the directional speed of a object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time

(i) Acceleration (a) of the particle at $t=3s$

$a=\frac{dv}{dt}$

$a=\frac{d}{dt}(9t^2+6t+3)$

$a=18+6$

$t=3s$

$a=(18\times 3)+6$

$=54+6$

$=60cm/s^2$

(ii) Given $v=9t^2+6t+3$

$\Rightarrow \frac{ds}{dt}\Rightarrow\frac{d}{dt}(9t^2+6t+3)$

$\Rightarrow ds=(3+6t+9t^2)dt$

On integration both sides, we get

$ds=\int\limits^8_5 {3+6t+9t^2} \,$

$\Rightarrow s=[3+\frac{6t^2}{2}+\frac{9t^3}{3}]^8_5$

$=1287m$

Hence, the acceleration a displacement $60cm/s^2$ and $1287m$.

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