Question

The displacement (in metre) of a particle moving along X axis is given by x=

3t2

+5t+5. Calculate:

(i) Instantaneous velocity at t=2s,

(ii) Average velocity between t=2s & t=4s,

(iii) Instantaneous acceleration at t=2s.​

Answer 1

Answer:

(i) Instantaneous velocity at t=2s = 17 m/s

(ii) Average velocity between t=2s & t=4s, = 23 m/s

(iii) Instantaneous acceleration at t=2s. = 6 m/s²

Explanation:

Given -

displacement = x = 3t²+5t+2

we know

(i) velocity (v) = dx/dt

(ii) acceleration (a )= dv/dt =d²x/dt²

| } x = 3t²+5t+2

differentiate both sides w.r.t t ( time)

dx/dt = 6t+5

v = 6t+5

-> at t = 2s

v = 6(2)+5

v = 17 m/s

|| } x = 3t²+5t+2

differentiate both sides w.r.t t ( time)

dx/dt = 6t+5

v = 6t+5

-> at t = 2s

v = 6(2)+5

v = 17

-> at t = 4s

v = 6(4)+5

v = 29

Average velocity = (17+29)/2 = 46/2 = 23

Average velocity ( t=2s & t=4s ) = 23 m/s

||| } x = 3t²+5t+2

differentiate both sides w.r.t t ( time)

dx/dt = 6t+5

v = 6t+5

again differentiate both sides w.r.t t ( time)

dv/dt = 6

a = 6 m/s²

Answer 2

Answer:

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Explanation:

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