Question

Ramu walks 3km/hr while his brother Somu walks at 4km/hr. They both started school, which is 2kms away from their house at 7:00 am. Then, which of these is correct?
a) Somu reaches school at 7:20 am
b) Somu reaches school 10 mins earlier than Ramu
c) Ramu reaches school at 7:30 am
d) All the above

Answer 1

Hint: Find the time taken by Ramu and Somu separately by using the formula of acceleration motion, which is an equation of motion. Find the time taken by both and compare it.

Complete step-by-step answer:

It is said that Ramu walks at the speed of 8km/hr. Similarly Somu walks at the rate of 4km/hr. Now both Ramu and Somu start walking at sharp 7:00 am.

The distance both Ramu and Somu have to walk to reach the school is 2km. Thus we know the distance travel and their speed, so we need to find the time taken by Somu and Ramu to reach school. Let us first take the case of Ramu. We know the formula of acceleration motion

$\begin{align}

  & \text{distance}=\text{velocity}\times \text{time}\Rightarrow s=vt \\

 & \therefore \text{time}=\dfrac{\text{distance}}{\text{velocity}}\Rightarrow t=\dfrac{s}{v} \\

\end{align}$

Speed of Ramu = 3km/hr, distance = 2km

Therefore time taken by Ramu

$=\dfrac{\text{distance travelled}}{\text{velocity}}$

Time taken by Ramu

$=\dfrac{2km}{3km/hr}=\dfrac{2}{3}\times 60=\dfrac{120}{3}=40\min $

Thus Ramu takes 40 mins to reach the school. Now, let us find the case of Somu
Time taken by Somu
$==\dfrac{\text{distance travelled}}{\text{velocity}}=\dfrac{2}{4}\times 60=\dfrac{120}{4}=30\min $

Therefore Somu takes 30 mins to reach the school. Thus Ramu takes 10 min more than Somu to reach the school at 7:30 am.

Thus Somu reaches school 10 mins earlier than Ramu.

So, option (b) is correct.

Note: s = vt is an equation of motion. To find the velocity, acceleration and to get the distance travelled, you need to know how to rearrange equations to make different quantities the subject of the equation.