Question

A car moves with a speed of \[40\text{ km/h}\] for \[\text{15}\] minutes and then with a speed of \[60\text{ km/h}\] for the next \[\text{15}\] minutes. The total distance covered by the car is:
A. \[35\]
B. \[25\]
C. \[45\]
D. \[66\]

Answer 1

Hint: The distance d covered by a body, moving with speed s, in time t is given by \[d=s\times t\]

Complete step by step solution:
Let \[{{d}_{1}}\] distance be covered in the first \[\text{15}\] minutes when the car travels with a speed \[{{s}_{1}}\] (say).
The speed of the car for the first \[\text{15}\] minutes is \[40\text{ km/h}\], so \[{{s}_{1}}=40\text{ km/h}\]
Convert the speed from km/h to metre per second using the conversion ratio \[1\text{ km/h}=\dfrac{1000\text{ }}{3600\text{ }}\text{m/s}\]
Therefore,
$
{{s}_{1}}=40\text{ km/h}=40\times \dfrac{1000\text{ }}{3600\text{ }}\text{ m/s} \\
{{s}_{1}}=11.11\text{ m/s} \\
$
Substituting \[{{s}_{1}}=11.11\text{ m/s}\] and \[{{t}_{1}}\text{=15 min}\times 60\text{ s}=900\text{ s}\] in the distance formula, the distance covered in first \[\text{15}\] minutes is
$
{{d}_{1}}={{s}_{1}}\times {{t}_{1}} \\
\implies {{d}_{1}}=11.11\text{ m/s}\times \text{900 s} \\
\implies {{d}_{1}}=9999\text{ m} \\
\implies {{d}_{1}}\cong 10\text{ km} \\
$

Let \[{{d}_{2}}\] distance be covered in the next \[\text{15}\] minutes when the car travels with a speed \[{{s}_{2}}\] (say).
Now, the speed of the car for the next \[\text{15}\] minutes is \[60\text{ km/h}\], so \[{{s}_{2}}=60\text{ km/h}\]
Convert the speed from km/h to metre per second using the conversion ratio \[1\text{ km/h}=\dfrac{1000\text{ }}{3600\text{ }}\text{m/s}\]
$
  {{s}_{2}}=60\text{ km/h}=60\times \dfrac{1000\text{ }}{3600\text{ }}\text{ m/s} \\
 {{s}_{2}}=16.67\text{ m/s} \\
$
Substituting \[{{s}_{2}}=16.67\text{ m/s}\] and \[{{t}_{2}}\text{=15 min}\times 60\text{ s}=900\text{ s}\] in the distance formula, the distance covered in next \[\text{15}\] minutes is
$
  {{d}_{2}}={{s}_{2}}\times {{t}_{2}} \\
\implies {{d}_{2}}=16.67\text{ m/s}\times \text{15 min} \\
\implies {{d}_{2}}=16.67\text{ m/s}\times \text{900 s} \\
\implies {{d}_{2}}=15003\text{ m} \\
\implies {{d}_{2}}\cong 15.0\text{ km} \\
$
Therefore total distance travelled by the car in \[\text{30}\] minutes is
$
  d={{d}_{1}}+{{d}_{2}} \\
\implies d=10\text{ km}+15\text{ km} \\
\therefore d=25\text{ km} \\
$

So, the total distance travelled by the car is \[25\text{ km}\].

Therefore, option B is the correct answer.

Additional information: As the speed increases after the first fifteen minutes, the car is accelerating.

Note: It is important that the distance, speed and time be in the same system of units, preferably the S.I. unit system.
The problem can also be solved by plotting a speed-time graph. Either convert speed into m/s and time into seconds, or convert only time into hours and plot the speed-time graph. Take the car to be initially at rest.