wiz-icon
MyQuestionIcon
MyQuestionIcon
10
You visited us 10 times! Enjoying our articles? Unlock Full Access!
Question

Acceleration of the particle in a rectilinear motion is given as a=4t+3 in (m/s2). The final position of the particle as a function of time t is
(Given, at t=0,u=3 m/s,x=2 m)

A
2t33+3t22+3t
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2t33+3t22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2t33+3t22+3t+2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2t33+3t223t
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2t33+3t22+3t+2
Given, a(t)=4t+3
As we know that the velocity as a fuction of time is given by
vudv=t0(4t+3)dt
[v]vu=[2t2+3t]t0
vu=2t2+3t
Or, v3=2t2+3t v(t)=2t2+3t+3
Now, final position of the particle is given as
xfxix(t)dt=t0v(t)dt=t0(2t2+3t+3)dt
xfxi=[2t33+3t22+3t]t0
xf2=[2t33+3t22+3t]
xf(t)=[2t33+3t22+3t+2]

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon