Question

Two cars are moving towards each other with velocity 4m/s and 8m/s respectively from two different points in a straight line, 1200m apart simultaneously, finding time after which both cars meet.
A. 300 sec
B. 150sec
C. 100sec
D. none of these

Answer 1

Hint:As a first step, one could read the question well and hence note down the values from the question. You could try making a rough diagram for better understanding. Then apply the formula for velocity for both cases and then find the answer to the question accordingly.

Formula used:
Velocity,
$v={\displaystyle \frac{S}{t}}$

Complete step-by-step solution:
In the question we are given the case where two cars, say $C_1$and$C_2$are moving towards each other with velocities 4m/s and 8m/s respectively along a straight line. At their respective initial positions, both were 1200m apart, we are supposed to find the time taken for the cars to meet.
Let us assume that the cars meet at a point at x cm distance from the first car$C_1$and it will be $(1200-x)cm$from the other car in time t.
Now, you may recall that the velocity is given by,
$v={\displaystyle \frac{S}{t}}$
For car $C_1$,
$v_1=4m{s}^{-1}={\displaystyle \frac{x}{t}}$
$\Rightarrow x=4t$……………………………………. (1)
Now for car $C_2$,
$v_2=8m{s}^{-1}={\displaystyle \frac{1200-x}{t}}$
$\Rightarrow 1200-x=8t$……………………………………. (2)
Takin the sum of equations (1) and (2),
$\Rightarrow 8t+4t=12t=1200$
$\therefore t=100\sec$
Therefore, we found the time taken by the cars to meet to be 100seconds.
Hence, option C is correct.

Note: The given question is a very basic one from the concept of one dimensional motion. Though the solution simply uses the basic algebra, we could solve this in an alternate way. Take the relative speed of the two cars, since they are traveling in opposite directions would be their sum. Divide the total distance by this relative speed to get the answer.
$t={\displaystyle \frac{S}{v_1+v_2}}={\displaystyle \frac{1200}{8+4}}=100s$