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Question: A car acquires a velocity of \[72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}\] in \[2\,{\text{s}}\] star...

A car acquires a velocity of 72kmh172\,{\text{km}} \cdot {{\text{h}}^{ - 1}} in 2s2\,{\text{s}} starting from rest. Calculate the average velocity.

A. 10ms110\,{\text{m}} \cdot {{\text{s}}^{ - 1}}

B. 5ms15\,{\text{m}} \cdot {{\text{s}}^{ - 1}}

C. 20ms120\,{\text{m}} \cdot {{\text{s}}^{ - 1}}

D. 72kmh172\,{\text{km}} \cdot {{\text{h}}^{ - 1}}

Explanation

Solution

Use the average velocity formula. Also determine the acceleration of the car and the total displacement of the car using kinematic equations relating the initial velocity, final velocity, acceleration, total displacement and time.

Formulae used:

The expression for the average velocity vv is

v=st\Rightarrow v = \dfrac{s}{t} …… (1)

Here, ss is the total displacement and tt is the time.

The kinematic equation relating final velocity vv, initial velocity uu, acceleration aa and time tt is

v=u+at\Rightarrow v = u + at …… (2)

The kinematic equation relating displacement ss, initial velocity uu, acceleration aa and time tt is

s=ut+12at2\Rightarrow s = ut + \dfrac{1}{2}a{t^2} …… (3)

Complete step by step answer:

The car starts from rest. Hence, the initial velocity of the car is zero.

The time required for the car to reach the velocity 72kmh172\,{\text{km}} \cdot {{\text{h}}^{ - 1}} is 2s2\,{\text{s}}.

Calculate the acceleration of the car.

Rearrange equation (2) for acceleration aa.

a=vut\Rightarrow a = \dfrac{{v - u}}{t}

Substitute 72kmh172\,{\text{km}} \cdot {{\text{h}}^{ - 1}} for vv, 0kmh10\,{\text{km}} \cdot {{\text{h}}^{ - 1}} for uu and 2s2\,{\text{s}} for tt in the above equation.

a=(72kmh1)(0kmh1)2s\Rightarrow a = \dfrac{{\left( {72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right) - \left( {0\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right)}}{{2\,{\text{s}}}}

a=(72kmh1)(103m1km)(1h3600s)2s \Rightarrow a = \dfrac{{\left( {72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\left( {\dfrac{{1\,{\text{h}}}}{{3600\,{\text{s}}}}} \right)}}{{2\,{\text{s}}}}

a=10ms2 \Rightarrow a = 10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}

Hence, the acceleration of the car is 10ms210\,{\text{m}} \cdot {{\text{s}}^{ - 2}}.

Calculate the total displacement of the car.

Substitute 0kmh10\,{\text{km}} \cdot {{\text{h}}^{ - 1}} for uu, 2s2\,{\text{s}} for tt and 10ms110\,{\text{m}} \cdot {{\text{s}}^{ - 1}} for aa in equation (3).

s=(0ms1)(2s)+12(10ms1)(2s)2\Rightarrow s = \left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)\left( {2\,{\text{s}}} \right) + \dfrac{1}{2}\left( {10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right){\left( {2\,{\text{s}}} \right)^2}

s=20m \Rightarrow s = 20\,{\text{m}}

Hence, the total displacement of the car is 20m20\,{\text{m}}.

Now calculate the average velocity of the car.

Substitute 20m20\,{\text{m}} for ss and 2s2\,{\text{s}} for tt in equation (1).

v=20m2s\Rightarrow v = \dfrac{{20\,{\text{m}}}}{{2\,{\text{s}}}}

v=10ms1 \Rightarrow v = 10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}

Therefore, the average velocity of the car is 10ms110\,{\text{m}} \cdot {{\text{s}}^{ - 1}}.

Hence, the correct option is A.

Note: The average velocity of the car can also be determined by taking the mean of the initial and final velocity of the car.Also don’t confuse between the average speed and average velocity,the former is defined as total distance travelled divided by the time taken while later is defined as the total displacement divided by total time taken.