Question 6: Motion in 1D

Question

Two cars $A$ and $B$ are moving in the same direction along a straight line with speeds $100 \mathrm{~km} / \mathrm{h}$ and $80 \mathrm{~km} / \mathrm{h}$, respectively such that car $A$ is moving ahead of car $B$. A person in car $B$ throws a stone with a speed $v$ so that it hits the car $A$ with a speed of $5 \mathrm{~m} / \mathrm{s}$. The value of $v$ is $\_\_\_\_$ $\mathrm{km} / \mathrm{h}$.

Accepted Answer

Take the direction of motion of the cars as positive.
Speeds of the cars are:
$$ v_A = 100 	ext{ km/h}, \qquad v_B = 80 	ext{ km/h} $$
A person in car $B$ throws a stone towards car $A$.
Let the speed of throw relative to car $B$ be $v$ km/h.
So, speed of the stone relative to ground will be:
$$ v_s = 80 + v 	ext{ km/h} $$
Now, the stone hits car $A$ with a speed of $5 	ext{ m/s}$ relative to car $A$.
Convert $5 	ext{ m/s}$ into km/h:
$$ 5 	imes \frac{18}{5} = 18 	ext{ km/h} $$
Since the stone must catch car $A$ from behind, its speed relative to car $A$ is:
$$ v_s - v_A = 18 $$
Substitute values:
$$ (80 + v) - 100 = 18 $$
$$ v - 20 = 18 $$
$$ v = 38 $$
Hence, the required value of $v$ is
$$ \boxed{38 	ext{ km/h}} $$
So, the correct option is
Option C
.

Answer

18

Answer

28

Answer

38

Answer

48

Motion in 1D

Choose the correct answer: