First slide

Graphical representation-Motion in 1 d

Question

The drawing shows velocity (v) versus time (r) graphs for two cyclists moving along the same straight segment of a highway from the same point. The second cyclist starts moving at t = 3 min. At what time do the two cyclists meet?

Difficult

A

4 min
B

6 min
C

8 min
D

12 min

Solution

$\begin{array}{l} \tan θ_{1} = \frac{v_{0}}{4}, \tan θ_{2} = \frac{v_{0}}{1} \\ \Rightarrow 4 \tan θ_{1} = \tan θ_{2} = v_{0} \end{array}$
Let they meet at time t, then their displacement should be same.
It means area under v-t graph should be same.
$\begin{array}{l} S_{1} = S_{2} \\ \Rightarrow \frac{1}{2} v_{1} t = \frac{1}{2} v_{2} (t - 3) \\ \Rightarrow \frac{1}{2} [(\tan θ_{1}) t] t = \frac{1}{2} [(\tan θ_{2}) (t - 3)] (t - 3) \\ \Rightarrow (\tan θ_{1}) t^{2} = 4 (\tan θ_{1}) (t - 3)^{2} \\ \Rightarrow t = 2 (t - 3) \Rightarrow t = 6 \min \end{array}$

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