Question

if the velocity of body v=2t^2-t, find the accleration at t=2 sec​

Answer 1

It is given that

a=

dt

dv

=

dt

d

(2t

2

e

−t

)

=2[t

2

e

−1

(−1)+e

−1

2t]

=e

−1

(4t−2t

2

)

So, a=0

⟹ 4t−2t

2

=0

Or 2t(2−t)=0

⟹ t=0 and t=2

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Answer 2

Given :

Velocity of the body is given by ,

• v = 2t² - t

To Find :

Acceleration at t = 2 s

Solution :

★ Acceleration : It is defined as change in velocity per change in time .

$\bullet\ \; \sf a=\dfrac{\Delta v}{\Delta t}$

At a particular instant of time ,

$\bullet\ \; \sf a=\lim _{\Delta t \to 0}\dfrac{dv}{dt}$

where ,

• a denotes instantaneous acceleration

____________________________________

★ v = 2t² - t

➠ a = $\sf \dfrac{dv}{dt}$

➠ a = $\sf \dfrac{d}{dt}(2t^2-t)$

➠ a = 2(2t) - (1)

➠ a = 4t - 1

Acceleration at t = 2 s is ,

➠ a = 4(2) - 1

➠ a = 8 - 1

➠ a = 7 m/s²