Question

At 9 am a car (A) began its journey from a point, travelling at 40mph. At 10am another car (B) started travelling from the same point at 60mph in the same direction as car (A). At what time will car (B) pass car (A)?

Answer 1

Hint: Assume the after time ‘t’ both the cars will be at the same position, i.e., they will meet. Consider \[{{d}_{A}}\] as the distance travelled by car A and \[{{d}_{B}}\] as the distance travelled by car B. Use the formula: - Speed \[\times \] time to find the values of \[{{d}_{A}}\] and \[{{d}_{B}}\] and equate the two distances to find the value of ‘t’. Use ‘t’ and (t – 1) as the time taken by car A and B respectively.

Complete step by step solution:
Here, we have been provided with the information regarding the journey of two cards with the speed of car B greater than that of car A. We have been asked to find the time at which car B will pass car A. So, basically we need to find the time at which they will meet at a particular point.
Now, let us assume that both the cars started from the point P and after a certain of time meet at point Q. So, we have,

Let us start counting the time from 9am and assume that after time ‘t’ both the cars meet. Since car A started the journey at 9am, it would have travelled ‘t’ hours. It is given that car B started the journey at 10am, so it would have travelled (t – 1) hours to meet at point Q.
Let, \[{{d}_{A}}\] = distance travelled by car A
Using the formula, speed \[\times \] time = distance, we have,
\[\Rightarrow {{d}_{A}}=40\times t\]
\[\Rightarrow {{d}_{A}}=40t\] ------ (1)
Also, \[{{d}_{B}}\] = distance travelled by car B
Using the formula, speed \[\times \] time = distance, we have,
\[\Rightarrow {{d}_{B}}=60\times \left( t-1 \right)\]
\[\Rightarrow {{d}_{B}}=60\left( t-1 \right)\] -------- (2)
Now, since both the cars have started from the same point and are meeting at the same point, so the distance travelled by them must be equal, so we have,
\[\Rightarrow {{d}_{A}}={{d}_{B}}\]
From equations (1) and (2), we get,
\[\Rightarrow 40t=60\left( t-1 \right)\]
Cancelling the common factors, we get,
\[\begin{align}
  & \Rightarrow 2t=3\left( t-1 \right) \\
 & \Rightarrow 2t=3t-3 \\
\end{align}\]
\[\Rightarrow t=3\] hours
Therefore, both the cars will meet after 3 hours, i.e. at 9 + 3 = 12pm.
Hence, 12pm is our answer.

Note: One may note that here we have taken the travelling time of car B as (t – 1) hours because it was given to us that car B started the journey 1 hours after car A started. So, to travel the same distance it required more speed. Remember that car B will pass car A just after the time they will meet so we have founded the time at which they will meet. You must remember the speed – time relation to solve the question.