Question

Find the instantaneous acceleration of a particle moving along x-axis with displacement x=18t + 5t^2​

Answer 1

Answer:

10m/s²

Explanation:

For instantaneous acceleration we have to double differentiate displacement.

x=18t +5t²

v=18+10t

a=10

Answer 2

Correct Question:- The displacement of a particle moving along x - axis is given by x = 18t + 5t² calculate the instantaneous velocity at t = 25. Average velocity at t = 25 and t = 35.

$\huge\bf\underline{\underline{\pink{A} \orange{N}\blue{S}\red{W}\green{E}\purple{R}}}$

Instantaneous velocity is given as time derivative of displacement which is V = DX/DT = 18 Unit.

Put T = 25, we get V = 268 Unit.

Average velocity,

$= \frac{net \: displacement}{time \: taken} =$

Position at time T = 0 is X = 0

Position at time T = 25 is X = 3575 Unit.

So displacement in time interval of

25 Second is 3575 - 0 = 3575

So Average velocity = 3575/25 = 143 Unit.

Position at T = 35S is X = 6755 Unit.

So Average velocity at T = 35S is 6755/35 = 193 Unit.