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Question: What average speed, most nearly, is required to run a mile (1.6 km) in 4 minutes? A. \[4.0m/s\] ...

What average speed, most nearly, is required to run a mile (1.6 km) in 4 minutes?
A. 4.0m/s4.0m/s
B. 7.0m/s7.0m/s
C. 40.0m/s40.0m/s
D. 400.0m/s400.0m/s

Explanation

Solution

Here, we will apply the basic relationship between speed, distance and time.For any moving object, speed is defined as the distance travelled per unit time. The unit of speed is calculated as the unit of distance divided by the unit of time.
Mathematically, the speed is calculated as,
Speed = DistanceTime{\text{Speed = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}}
The SI unit of speed is ms\dfrac{{\text{m}}}{{\text{s}}} and its dimension is, [LT - 1]\left[ {{\text{L}}{{\text{T}}^{{\text{ - 1}}}}} \right]

Complete step-by-step solution:
Given that:
Distance is given as,

d=1mile =1.6km  d = 1{\text{mile}} \\\ = 1.6{\text{km}} \\\

Travelling time is, t=4minutest = 4{\text{minutes}}
From the basic conversion of primary units, we know that1km=1000m1{\text{km}} = 1000{\text{m}}
So the travelling distance will be converted as, d=1.6×103md = 1.6 \times {10^3}{\text{m}}.
From the basic conversion of primary units, we also know that1minute=60seconds1{\text{minute}} = 60{\text{seconds}}
So the travelling time can be converted as, t=4×60=240st = 4 \times 60 = 240{\text{s}}.
To solve this problem,we apply the basic relationship between velocity(speed),distance and time.
i.e. Speed = DistanceTime{\text{Speed = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}}
Now put all the given values in their required converted units.

Speed = DistanceTime =1.6×103m240s =6.67ms  {\text{Speed = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}} \\\ = \dfrac{{1.6 \times {{10}^3}{\text{m}}}}{{240{\text{s}}}} \\\ = 6.67\dfrac{{\text{m}}}{{\text{s}}} \\\

In this problem, the most nearly value of average speed is asked to calculate.The most nearly value of 6.67ms6.67\dfrac{{\text{m}}}{{\text{s}}} will be 7.0ms.7.0\dfrac{{\text{m}}}{{\text{s}}}.
Therefore, the most nearly value of average speed is calculated as, 7.0ms.7.0\dfrac{{\text{m}}}{{\text{s}}}.
Hence, the correct option for this problem is, B.

Note:- First of all, in this problem students can get confused between the terms speed and velocity. We need to understand that speed is a scalar quantity (it does not depend on the direction or position of the object). Mathematically, Speed = DistanceTime{\text{Speed = }}\dfrac{{{\text{Distance}}}}{{{\text{Time}}}}
While, velocity is a vector quantity (completely depends on the direction or position of the object).
Mathematically, Velocity = DisplacementTime{\text{Velocity = }}\dfrac{{{\text{Displacement}}}}{{{\text{Time}}}}