At t=0, a particle starts from (1,0) and moves towards positive x−axis with speed of v=3t2+2tm/s. The final position of the particle as a function of time is
A
3t32−2t2+1m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3t32+2t2+1m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
t3+t2+1m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
t3−t2+1m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ct3+t2+1m Given, particle starts from (1,0), so the initial position of the particle is xi=1
As we know that, ∫dx=∫t0v(t)dt ⇒∫xfxidx=∫t0(3t2+2t)dt ⇒(xf−xi)=[t3+t2]t0 ⇒xf−1=t3+t2⇒xf=t3+t2+1m