Question

the displacement of a particle at any instant is given by 4t2+8t-4 find the velocity when t = 3

Answer 1

GiveN :

• x = 4t² + 8t - 4

To FinD :

• Velocity when t = 3

SolutioN :

As we know that :

$\dashrightarrow {\boxed{\tt{v \: = \: \dfrac{dx}{dt}}}} \\ \\ \dashrightarrow \tt{v \: = \: \dfrac{d(4t^2 \: + \: 8t \: - \: 4)}{dt}} \\ \\ \footnotesize {\underline{\sf{\dag \: \: \: \: \: Differentiate \: x \: wrt. \: dt \: \: \: \: \: \:}}} \\ \\ \dashrightarrow \tt{v \: = \: 4(2t) \: + \: 8 \: - \: 0} \\ \\ \dashrightarrow \tt{v \: = \: 8t \: + \: 8} \\ \\ \large{\boxed{\boxed{\sf{Velocity \: = \: 8t \: + \: 8 \: ms^{-1}}}}}$

____________________________

We have to find velocity in 3s. So, put t = 3

$\dashrightarrow \tt{v \: = \: 8(3) \: + \: 8} \\ \\ \dashrightarrow \tt{v \: = \: 24 \: + \: 8} \\ \\ \dashrightarrow \tt{v \: = \: 32} \\ \\ \underline{\sf{\therefore \: Velocity \: when \: time \: is \: 3s \: is \: 32 \: ms^{-1}}}$

Answer 2

Answer:

Given:

Displacement function has been provided as follows :

s = 4t² + 8t - 4

To find:

Velocity at t = 3 sec.

Calculation:

First of all, we need to differentiate the displacement function with respect to time. It would give us the instantaneous velocity . Then if we put the value the value of t = 3 sec , we get tge instantaneous Velocity at 3 sec .

s = 4t² + 8t - 4

=> v = ds/dt

=> v = d(4t² + 8t - 4)/dt

=> v = 8t + 8

Putting t = 3 sec , we get :

=> v = (8×3) + 8

=> v = 24 + 8

=> v = 32 m/s

So final answer :

$\boxed{ \red{ \huge{ \bold{v = 32 \: m {s}^{ - 1} }}}}$