A boy is running at a speed of $p$ km/hr to cover a distance of 1km. But, due to slippery ground, his speed is reduced by $q$ km/hr $\left( p>q \right)$. If he takes $r$ hours to cover the distance, then,
(a) $\dfrac{1}{r}=\left( p-q \right)$
(b) \[r=\left( p-q \right)\]
(c) \[\dfrac{1}{r}=\left( p+q \right)\]
(d) \[r=\left( p+q \right)\]
Answer
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Hint: To solve this problem, the formula that has to be remembered is $\text{Speed=}\dfrac{\text{Distance}}{\text{Time taken}}$ . Where distance is in kilometres and the time is in hours. Hence the speed is in the units km\hr. It is said in the question that the boy’s speed is reduced, so we can get the reduced speed as (p-q). Now, we have the speed and distance is 1km. We can now find the time in $r$ hours using the above-mentioned formula.
Complete step by step solution:
As it is given in the question that the boy is running at a speed of $p$ km/hr, therefore there is no acceleration involved. So, acceleration is considered to be zero throughout the problem.
We can see that earlier the boy was running at a speed of $p$ km/hr and the speed is reduced by $q$ km/hr.
Therefore, the new speed is given as follows,
New speed = Old speed – speed by which is it reduced.
So, from the condition, we obtained the following relation,
Reduced speed $=\left( p-q \right)$km/hr…………….(i)
This is the new speed by which the boy runs and covers the distance of 1 km.
The total time taken to complete the travel is given by $r$hours……………….(ii)
We know that the relation of the speed, distance and the time taken is given by,
$\text{Speed=}\dfrac{\text{Distance}}{\text{Time taken}}........(iii)$
Therefore from (i),(ii),(iii) we get,
$\left( p-q \right)=\dfrac{1}{r}........................(v)$
By rearranging the terms we get,
$\dfrac{1}{r}=\left( p-q \right)$
Hence, this is the required solution.
So, the correct option is (a).
Note: Here, the units on L.H.S and R.H.S must be the same. Using this, we can even eliminate options b and d where the unit on L.H.S. is in hours and on R.H.S. it is in km/hr. Then we will be left with two options and can choose the right one with the help of formula. Also, the condition (p > q) has to be taken into notice. The condition says that the direction of the motion of the boy has not been changed. In short, the reduced speed of the boy is less the speed of the boys without the slippery road.
Complete step by step solution:
As it is given in the question that the boy is running at a speed of $p$ km/hr, therefore there is no acceleration involved. So, acceleration is considered to be zero throughout the problem.
We can see that earlier the boy was running at a speed of $p$ km/hr and the speed is reduced by $q$ km/hr.
Therefore, the new speed is given as follows,
New speed = Old speed – speed by which is it reduced.
So, from the condition, we obtained the following relation,
Reduced speed $=\left( p-q \right)$km/hr…………….(i)
This is the new speed by which the boy runs and covers the distance of 1 km.
The total time taken to complete the travel is given by $r$hours……………….(ii)
We know that the relation of the speed, distance and the time taken is given by,
$\text{Speed=}\dfrac{\text{Distance}}{\text{Time taken}}........(iii)$
Therefore from (i),(ii),(iii) we get,
$\left( p-q \right)=\dfrac{1}{r}........................(v)$
By rearranging the terms we get,
$\dfrac{1}{r}=\left( p-q \right)$
Hence, this is the required solution.
So, the correct option is (a).
Note: Here, the units on L.H.S and R.H.S must be the same. Using this, we can even eliminate options b and d where the unit on L.H.S. is in hours and on R.H.S. it is in km/hr. Then we will be left with two options and can choose the right one with the help of formula. Also, the condition (p > q) has to be taken into notice. The condition says that the direction of the motion of the boy has not been changed. In short, the reduced speed of the boy is less the speed of the boys without the slippery road.
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