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A particle travels in a straight line such that for a short time 2s t 6s, its motion is described by v=(4/a)m/s, where 'a' is in m/s 2 . If v=6m/s when t=2sec, determine the particle's acceleration when t=3 sec.

A particle travels in a straight line such that for a short time 2s<t<6s, its motion is described by v=(4/a)m/s, where 'a' is in m/s2. If v=6m/s when t=2sec, determine the particle's acceleration when t=3 sec.

Grade:12

2 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago

Dear Chilukuri

V=4/a

Va =4

V dv/dt =4

intigrate

V2/2 =4t +c1

V2 =8t +c 

apply given condition

36 =16 +c

c= 20

so

V2 =8t +20 

velocity at t=3 sec

V = √44 

so acceleration  a=4/v  =4/√44

                                =2/√11 m/sec2

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Badiuddin

LISA
15 Points
9 months ago
From the given data:
v=4/ams−1
Acceleration From on=? At time t=3s
Given velocity is
 v=4/a⟹a=4/v⟹dv/dt=4/v.......(1) 
Let us integrate on both the side
 ∫vdv=∫4dt⟹v²/2−18=4t−8 
 v²/2=4t−8+18⟹v²=8t+20⟹v=√8t+20
Equation 1 implies
 a=dv/dt⟹a=√8t+20
The acceleration of the particle at time t = 3s
 a=4/√8(3)+20⟹a=0.603ms−²

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