Question

HEYA........ KARTIK HERE

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◼️NEED FULL EXPLANATION
◼️10 POINTS
◼️NO SPAMS

Answer 1

•according to the question we should calculate the position of the particle .

1 ) given :

S =t^3 - 2t^2 + 5t + 4

At , t = 1 sec the position of the particle is

S = ( 1 ) ^3 - 2 (1)^2 + 5 (1) + 4

S = 1 - 2 +5 +4

S = 8 .

2) at first derivative

dS/dt = 3t^2 -4t + 5

Given t = 1 sec

Therefore,

dS/dt = 3(1)^2 -4(1) +4

dS/dt = 3

Implies at first derivative, S = 3

3) at second derivative

i) we know that the first derivative is

dS/dt = 3t^2 - 4t + 5

The second derivative is

d^2S / dt^2 = 6t -4

Given t = 1 sec

Therefore

d^2S/dt = 6(1) -4

= 2

Therefore the S = 2 at second derivative.

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Answer 2

Answer:

thanks for the free points