Question

If position of particles is x = ( 3t² - 2t + 1 ) m. find :- 1) velocity at time t 2) a acceleration at time t ?​

Answer 1

Step-by-step explanation:

Given that :

Position of particles (x) = 3t^2 - 2t +1

To Find :

• Velocity
• Acceleration

Solution :

As we know,

Velocity(V) = dx/dt

=> V = d(3t^2 - 2t + 1)/dt

=> V = 3×2t^(2 - 1) - 2×1t^(1-1) +0

----------------------¦

=> V = 6 t - 2 ¦

----------------------¦

Acceleration (a) = dv/dt

=> a = d(6t - 2)/dt

=> a = 6×1 t^1-1 - 0

=> a = 6 t^0

=> a = 6 ×1

=> a = 6 m/s Ans.

Answer 2

Answer:

VELOCITY = dx/dt

V = d (3t² - 2t +1 ) / dt

V = 3 ×2t ^(2-1) - 2×1t ^ (1-1) +0

V = 6t -2

ACCELERATION = dv / dt

A = d(6t -2) / dt

A = 6 ×1 t^1-1 - 0

A = 6t^ 0

A = 6 × 1

A = 6m/s

HOPE IT HELPS YOU OUT

PLS MARK AS BRAINLIEST ANSWER

FOLLOW ME TOO FOR MORE ANSWERS