A body starts from rest with a uniform acceleration. If its velocity after n second is v, then its displacement in the last 2s is
A
2v(n+1)n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
v(n+1)n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
v(n−1)n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2v(n−1)n
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D2v(n−1)n By using first equation of motion v=0+na(∵u=0) ⇒a=vn Displacement covered in n seconds is ⇒Sn=12an2 and distance travelled in (n−2) seconds is Sn−2=12a(n−2)2 So, displacement travelled in the last 2s is Sn−Sn−2=12an2−12a(n−2)2 =a2[n2−(n−2)2] Sn−Sn−2=a2(n2−n2−4+4n)=2a(n−1) Sn−Sn−2=2v(n−1)n