Question

-The velocity of a particle is given by
v=2t²-3t+10 m/s. Find the instantaneous
acceleration at t=5s.
​

Answer 1

Given

Velocity of the particle is given by v = 2t² - 3t + 10

To Find

Instantaneous acceleration at t = 5 s

Knowledge Required

Acceleration at a particular time is called instantaneous acceleration

While solving these type of questions , we just need to differentiate the velocity .

Since , acceleration is defined as rate of change in velocity .

$\bf \pink{\bigstar\ \; a=\dfrac{dv}{dt}}$

Solution

v = 2t² - 3t + 10

$\implies \rm a=\dfrac{dv}{dt}\\\\\implies \rm a=\dfrac{d}{dt}(2t^2-3t+10)\\\\\implies \rm a=4t-3$

Now , we need to find instantaneous acceleration at t = 5 s .

$\implies \rm a=4(5)-3\\\\\implies \rm a=20-3\\\\\implies \bf \green{a=17\ m/s^2\ \; \bigstar}$

Answer 2

$\mathcal{\gray{\underline{\underline{\blue{GIVEN:-}}}}}$

• The velocity of a particle given by v = “2t² - 3t + 10” m/s .

$\mathcal{\gray{\underline{\underline{\blue{TO\:FIND:-}}}}}$

• The instantaneous acceleration at 5s .

$\mathcal{\gray{\underline{\underline{\blue{SOLUTION:-}}}}}$

We have know that,

$\green\bigstar\:\rm{\purple{\boxed{\red{Acceleration\:=\:\dfrac{Velocity}{Time}\:}}}}$

✍️ To find the instantaneous acceleration, we can calculate the differentiation of velocity function with respect to time .

$\green\bigstar\:\rm{\purple{\boxed{\red{Instantaneous\:Acceleration\:=\:\dfrac{d(Velocity)}{d(Time)}\:}}}}$

$\purple\bigstar\:\rm{\gray{\overbrace{\underbrace{\orange{Instantaneous\:acceleration\:=\:\dfrac{dv}{dt}\:}}}}}$

$\rm{\implies\:Acceleration_{(insta)}\:=\:\dfrac{d(2t^2\:-\:3t\:+\:10)}{dt}\:}$

$\rm{\implies\:Acceleration_{(insta)}\:=\:2\dfrac{d(t^2)}{dt}\:-\:3\dfrac{d(t)}{dt}\:+\:10\dfrac{d(1)}{dt}\:}$

• d(t²)/dt = 2t

• d(t)/dt = 1

• d(1)/dt = 0

$\rm{\implies\:Acceleration_{(insta)}\:=\:2\times{2t}\:-\:3\times{1}\:+\:10\times{0}\:}$

$\rm{\implies\:Acceleration_{(insta)}\:=\:4t\:-\:3\:+\:0\:}$

$\rm{\green{\implies\:Acceleration_{(insta)}\:=\:4t\:-\:3}}$

⚡ Now put the value of “t = 5s” in the above equation

$\rm{\implies\:Acceleration_{(insta)}\:=\:4\times{5}\:-\:3}$

$\rm{\implies\:Acceleration_{(insta)}\:=\:20\:-\:3}$

$\rm{\pink{\implies\:Acceleration_{(insta)}\:=\:17\:m/s^2\:}}$

$\rm{\red{\therefore}}$ The instantaneous acceleration at 5s is “17 m/s²” .