
How long will a boy take to run around a square field of side 35 meters, if he runs at the rate of 9km/hr?
Answer
571.8k+ views
Hint: In order to solve this problem first we will calculate the perimeter of the square it means we get the total distance covered by a boy further we will evaluate the time taken to cover such a distance by dividing the total distance covered by boy by the speed of boy.
Complete step-by-step solution:
Given that side of a square field \[ = 35m\]
Distance runaround the field \[(d)\] = Perimeter of the field
As we know that perimeter of square is given as
Perimeter of square \[ = 4 \times \]side of square
So, \[d\]\[ = 4 \times 35m = 140m\]
It is given that speed of boy = 9km/hr
We will convert its unit with m/s, so we get
\[9km/hr = (9 \times \dfrac{5}{{18}})m/s = \dfrac{5}{2}m/s\]
Therefore, time taken to run around the field = Distance covered by boy \[/\] speed of boy
By substituting the values in the above formula, we get
\[
= \dfrac{{140m}}{{\dfrac{5}{2}m/s}} \\
= 140 \times \dfrac{2}{5}s \\
= 28 \times 2s \\
= 56s \]
Hence, time is taken to run around the field \[ = 56\]seconds.
Note: In order to solve for the distance, we used the formula $x = vt$ (where x represents the distance, v represents the speed and t represents the time taken to cover distance) or we can say that distance equals speed times time. Rate and speed are similar in this problem since they both represent some distance per unit time like miles per hour or kilometers per hour.
Complete step-by-step solution:
Given that side of a square field \[ = 35m\]
Distance runaround the field \[(d)\] = Perimeter of the field
As we know that perimeter of square is given as
Perimeter of square \[ = 4 \times \]side of square
So, \[d\]\[ = 4 \times 35m = 140m\]
It is given that speed of boy = 9km/hr
We will convert its unit with m/s, so we get
\[9km/hr = (9 \times \dfrac{5}{{18}})m/s = \dfrac{5}{2}m/s\]
Therefore, time taken to run around the field = Distance covered by boy \[/\] speed of boy
By substituting the values in the above formula, we get
\[
= \dfrac{{140m}}{{\dfrac{5}{2}m/s}} \\
= 140 \times \dfrac{2}{5}s \\
= 28 \times 2s \\
= 56s \]
Hence, time is taken to run around the field \[ = 56\]seconds.
Note: In order to solve for the distance, we used the formula $x = vt$ (where x represents the distance, v represents the speed and t represents the time taken to cover distance) or we can say that distance equals speed times time. Rate and speed are similar in this problem since they both represent some distance per unit time like miles per hour or kilometers per hour.
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