Question 136: Motion in 1D

Question

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 xQ(t) = (ft $$-$$ t2). 
At what time do the cars have the same velocity ?

Accepted Answer

For car P,
x
P
(t) =
(at + bt
2
)
v
P
(t) = $${{d{x_p}\left( t ight)} \over {dt}}$$ =
a + 2bt
Similarly for car Q,
x
Q
(t) =
(ft - t
2
)
v
Q
(t) = $${{d{x_Q}\left( t ight)} \over {dt}}$$ =
f - 2t
When they have same velocity then, v
P
(t) = v
Q
(t)
$$ 	herefore $$
a + 2bt = f - 2t
$$ \Rightarrow $$
2t(b + 1) = f - a
$$ \Rightarrow $$ t = $${{f - a} \over {2\left( {1 + b} ight)}}$$

Answer

$${{a - f} \over {1 + b}}$$

Answer

$${{a + f} \over {2\left( {b - 1} \right)}}$$

Answer

$${{a + f} \over {2\left( {1 + b} \right)}}$$

Answer

$${{f - a} \over {2\left( {1 + b} \right)}}$$

Motion in 1D

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 xQ(t) = (ft $-$ t2). At what time do the cars have the same velocity ?

Choose the correct answer: