Two cars P and Q start from a point at the same
time in a straight line and their positions are
represented by xp(t) = at + bt2 and xo(t) = ft – t2.
At what time do the cars have the same velocity?
Answers
Position of car P x
p
(t)=at+bt
2
Thus velocity of car P v
p
=
dt
d[x
p
(t)]
=a+2bt
Position of car Q x
Q
(t)=ft−t
2
Thus velocity of car Q v
Q
=
dt
d[x
Q
(t)]
=f−2t
According to the question, v
p
∣
∣
∣
∣
∣
t=t
o
= v
Q
∣
∣
∣
∣
∣
t=t
o
∴ a+2bt
o
=f−2t
o
⟹t
o
=
2(b+1)
f−a
At time
Answer:
t = \dfrac{f-a}{2(1+b)}
the cars will have same velocity
Given that,
Position of P =
X_{p}(t)=at+bt^2
Position of Q=
X_{q}(t)=ft+ft^2
We know that,
Velocity of P=
v _{P}=\dfrac{d(X_{P}(t))}{dt}
Velocity of Q=
v _{P}=a+2bt
v _{Q}=\dfrac{d(X_{Q}(t))}{dt}
v _{Q}=f+2ft
When the cars have same velocity,
v_{P}=v_{Q}
a+2bt=f+ft
The time will be
t = \dfrac{f-a}{2(1+b)}
Hence, At this time the cars will have same velocity .