Question

If position of particle is x = (t³-2t+1) find :- 1) velocity at t = 2sec. 2) acceleration at t = 2 sec ???​

Answer 1

$\huge \red Solution:-$

Velocity of the particle = t² + 2t + 2

dx/dt = t² + 2t + 2

⇒ dx = (t² + 2t + 2)dt.

Integrating both the sides of the equation.

x = t³/3  + t² + 2t + c

where c is the constant of the Integration.

Solving for c,

at time t = 0 the displacement is 0,

∴ 2 = 0 + 0 + 0 +  c

∴   ║ c = 2 ║

∴ Position of the Particle is given by,

x =  t³/3  + t² + 2t + 2

⇒ x = (2)³/3 + 4 + 2 + 4

⇒ x = 8/3 + 10

∴ x = (8 + 30)/3

∴ x = 38/3 m.

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Answer 2

Answer:

hey mate ✌✌

1) velocity will be 10m/s.

2) acceleration will be 12 m/s² .

the following answer is in attachment........

Explanation:

hope it helps ✔✔