Question

the displacement 's' of a moving particle at time 't' is given by s=5 + 20t - 2t^2. Find its velocity and acceleration when t = 2​

Answer 1

$\underline{\underline{\sf{\color{orange}{GIVEN:-}}}}$

The displacement 's' of a moving particle at time 't' is given by s = 5 + 20t - 2t².

$\underline{\underline{\sf{\color{orange}{TO\ FIND:-}}}}$

The velocity and acceleration of the moving particle at t = 2 s.

$\underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}$

$\sf{s=5+20t-2t^2}$

$\sf{Velocity,\ v=\dfrac{ds}{dt} }$

$\sf{\longmapsto v=\dfrac{d(5+20t-2t^2)}{dt} }$

$\sf{\longmapsto v=0+20+(2\times-2t)}$

$\sf{\longmapsto v=20-4t}$            . . . (i)

_______________________

Using equation (i) :-

$\sf{\longmapsto a=\dfrac{dv}{dt} }$

$\sf{\longmapsto a=\dfrac{d(20-4t)}{dt} }$

$\sf{\longmapsto a=0-4}$

$\sf{\longmapsto a=-4\ m/s^2}$

_______________________

Putting t = 2,

$\sf{\longmapsto v=20-4(2)}$

$\sf{\longmapsto v=20-8}$

$\sf{\longmapsto v=12\ m/s}$

Therefore,

◙ The velocity of the moving particle at t = 2s is 12 m/s.

◙ The acceleration of the moving particle at t = 2s = -4 m/s².

Answer 2

Given :

• Displacement of a moving particle at time 't' is given by, s (t) = 5 + 20 t - 2 t²

To find :

• Velocity and acceleration of particle when t = 2

Knowledge required :

• Formula to calculate instantaneous velocity

$\red{\bigstar\;\;\;}\boxed{\sf{v=\dfrac{d\;s}{d\;t}}}$

[ Where the quantity v (instantaneous velocity) is the differencial coefficient of x with respect to t and is denoted by (ds/dt) ]

• Formula to calculate instantaneous Acceleration

$\red{\bigstar\;\;\;}\boxed{\sf{a=\dfrac{d\;v}{d\;t}}}$

[ Where the quantity a (instantaneous acceleration) is the differential coefficient of v with respect to t and is denoted by (dv/dt) ]

Solution :

Using formula to calculate instantaneous velocity

$\implies \sf{v=\dfrac{d\;s}{d\;t}}$

$\implies \sf{v=\dfrac{d\;( -2\;t^2 + 20 \;t + 5)}{d\;t}}$

$\implies \underline{\underline{\red{\sf{v=-4\;t + 20 }}}}$

Using formula to calculate instantaneous acceleration

$\implies\sf{a=\dfrac{d\;v}{d\;t}}$

$\implies\sf{a=\dfrac{d\;(-4\;t+20)}{d\;t}}$

$\implies\underline{\underline{\red{\sf{a=-4}}}}$

So,

velocity at t = 2 would be

$\implies\sf{v (t)=-4\;t+20}$

$\implies\sf{v (2)=-4\;\times 2+20}$

$\implies\underline{\underline{\red{\sf{v (2)=12}}}}$

• velocity of particle at t = 2  would be 12.

and, acceelation is a constant as , a = -4

therefore,

• Acceleration of particle at t = 2 sec would be -4.