Question

1
(i) A particle is moving along a line such that s = 1/3t^2+ 8t + 5 (s is in ft., t is in sec.).
) s
Find its velocity and acceleration at t =2sec
​

Answer 1

Answer:

28/3 ft/s & 2/3 ft/s²

Step-by-step explanation:

dS/dt = dv

=> d(1/3t² + 8t + 5)/dt = dv

=> (1/3)*2*t + 8(1) + 0 = dv

=> (2/3)t + 8 = dv

At t = 2, v = (2/3)2 + 8 = 4/3 + 8 = 28/3 ft/s

Also,

dv/dt = da

=> d((2/3)t + 8)/dt = da

=> (2/3) + 0 = da

=> 2/3 ft/s² = da = acceleration.

Answer 2

GiveN :

s = 3t² + 2t + 1

To FinD :

Velocity of Particle at t = 2 s

SolutioN :

⇒s = 3t² + 2t + 1

Diff wrt dt

⇒ds/dt = d(3t² + 2t + 1)/dt

⇒v = 2*3t + 2 + 0

⇒v = 6t + 2__________[1]

ㅤㅤㅤㅤㅤㅤㅤPut t = 2 s

⇒v = 6(2) + 2

⇒v = 12 + 2

⇒v = 14

\therefore∴ Velocity at t = 2s is 14 m/s

_________________________

Diff (1) wrt. dt

⇒dv/dt = d(6t + 2)dt

⇒a = 6 + 0

⇒a = 6

\therefore∴ Acceleration of particle is 6 m/s²