1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Two cars P and Q starts from a point at the same time in a straight line and their positions are represented by xp(t)=at+bt2 and xQ(t)=ftâˆ’t2. At what time do the cars have the same velocity?

A
fa2(1+b)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
af1+b
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
a+f2(b1)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D

a+f2(1+b)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A f−a2(1+b)Position of car P, xp(t)=at+bt2 Thus, velocity of car P vp=d[xp(t)]dt=a+2bt Position of car Q xQ(t)=ft−t2 Thus, velocity of car Q vQ=d[xQ(t)]dt=f−2t According to the question, vp∣∣∣t=to= vQ∣∣∣t=to ∴a+2bto=f−2to ⟹to=f−a2(1+b)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Speed and Velocity
PHYSICS
Watch in App
Join BYJU'S Learning Program