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# A car acquires a velocity of $72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ in $2\,{\text{s}}$ starting from rest. Calculate the average velocity.A. $10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$B. $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$C. $20\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$D. $72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$

Last updated date: 20th Jun 2024
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Hint: 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 $v$ is
$\Rightarrow v = \dfrac{s}{t}$                                         …… (1)
Here, $s$  is the total displacement and $t$ is the time.
The kinematic equation relating final velocity $v$, initial velocity $u$, acceleration $a$ and time $t$ is
$\Rightarrow v = u + at$                                  …… (2)
The kinematic equation relating displacement $s$, initial velocity $u$, acceleration $a$ and time $t$ is
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$            …… (3)

The car starts from rest. Hence, the initial velocity of the car is zero.
The time required for the car to reach the velocity $72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ is $2\,{\text{s}}$.
Calculate the acceleration of the car.
Rearrange equation (2) for acceleration $a$.
$\Rightarrow a = \dfrac{{v - u}}{t}$
Substitute $72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ for $v$, $0\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ for $u$ and $2\,{\text{s}}$ for $t$ in the above equation.
$\Rightarrow a = \dfrac{{\left( {72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right) - \left( {0\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right)}}{{2\,{\text{s}}}}$
$\Rightarrow a = \dfrac{{\left( {72\,{\text{km}} \cdot {{\text{h}}^{ - 1}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\left( {\dfrac{{1\,{\text{h}}}}{{3600\,{\text{s}}}}} \right)}}{{2\,{\text{s}}}}$
$\Rightarrow a = 10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$
Hence, the acceleration of the car is $10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$.
Calculate the total displacement of the car.
Substitute $0\,{\text{km}} \cdot {{\text{h}}^{ - 1}}$ for $u$, $2\,{\text{s}}$ for $t$ and $10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for $a$ in equation (3).
$\Rightarrow s = \left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)\left( {2\,{\text{s}}} \right) + \dfrac{1}{2}\left( {10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right){\left( {2\,{\text{s}}} \right)^2}$
$\Rightarrow s = 20\,{\text{m}}$
Hence, the total displacement of the car is $20\,{\text{m}}$.
Now calculate the average velocity of the car.
Substitute $20\,{\text{m}}$ for $s$ and $2\,{\text{s}}$ for $t$ in equation (1).
$\Rightarrow v = \dfrac{{20\,{\text{m}}}}{{2\,{\text{s}}}}$
$\Rightarrow v = 10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
Therefore, the average velocity of the car is $10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.
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.