
A car requires $30$ seconds to accelerate from $0$ to $90km/hr$, its average acceleration is, most nearly.
A) $800m/{s^2}$
B) $80m/{s^2}$
C) $8m/{s^2}$
D) $0.8m/{s^2}$
E) $0.08m/{s^2}$
Answer
549.9k+ views
Hint: First we need to change the value of velocity from $km/hr$ to ${{m/s}}$. The average of any quantity is the difference between its initial value and the final value divided by the time taken to achieve the final value from the initial state.
Formula used:
$a = \dfrac{{\Delta V}}{{\Delta t}} = \dfrac{{{V_f} - {V_i}}}{{\Delta t}}$
where ${{{V}}_i}$is the initial velocity,
${{{V}}_f}$is the final velocity,
${{t}}$ is the time taken for the velocity to change from its initial value to its final value.
Complete step by step solution:
Acceleration of a body is the rate of change of velocity of the body. Average acceleration refers to the rate at which the velocity of an object changes. To calculate average acceleration, we need to divide the change in velocity by the time taken. The S.I unit of acceleration is ${{m/}}{{{s}}^2}$.
Average acceleration $a = \dfrac{{\Delta V}}{{\Delta t}} = \dfrac{{{V_f} - {V_i}}}{{\Delta t}}$,
where ${{{V}}_i}$is the initial velocity,
${{{V}}_f}$is the final velocity and
$\Delta t$ is the time taken for the velocity to change from ${{{V}}_i}$ to ${{{V}}_f}$.
But first, we need to change the unit of velocity from $km/hr$ to ${{m/s}}$.
As we know that
$1km/hr$ = $0.278m/s$
$90km/hr = 0.278 \times 90m/s = 25.02m/s$
As given in the question, the initial velocity of the car $ = 0$
Therefore, the average acceleration
$\Rightarrow a = \dfrac{{25.02 - 0}}{{30}} = 0.834{{m/}}{{{s}}^2}$
Therefore, the average acceleration of the car is $0.834m/{s^2}$.
The correct option is (D), $0.8m/{s^2}$.
Note: We need to take care that the velocity is changed from $km/hr$ to ${{m/s}}$ because the time is given in seconds and the S.I unit of acceleration is ${{m/}}{{{s}}^2}$. The question asks the value of acceleration that is most near to the given options, therefore we rounded off the value of the calculated average acceleration to one-decimal place.
Formula used:
$a = \dfrac{{\Delta V}}{{\Delta t}} = \dfrac{{{V_f} - {V_i}}}{{\Delta t}}$
where ${{{V}}_i}$is the initial velocity,
${{{V}}_f}$is the final velocity,
${{t}}$ is the time taken for the velocity to change from its initial value to its final value.
Complete step by step solution:
Acceleration of a body is the rate of change of velocity of the body. Average acceleration refers to the rate at which the velocity of an object changes. To calculate average acceleration, we need to divide the change in velocity by the time taken. The S.I unit of acceleration is ${{m/}}{{{s}}^2}$.
Average acceleration $a = \dfrac{{\Delta V}}{{\Delta t}} = \dfrac{{{V_f} - {V_i}}}{{\Delta t}}$,
where ${{{V}}_i}$is the initial velocity,
${{{V}}_f}$is the final velocity and
$\Delta t$ is the time taken for the velocity to change from ${{{V}}_i}$ to ${{{V}}_f}$.
But first, we need to change the unit of velocity from $km/hr$ to ${{m/s}}$.
As we know that
$1km/hr$ = $0.278m/s$
$90km/hr = 0.278 \times 90m/s = 25.02m/s$
As given in the question, the initial velocity of the car $ = 0$
Therefore, the average acceleration
$\Rightarrow a = \dfrac{{25.02 - 0}}{{30}} = 0.834{{m/}}{{{s}}^2}$
Therefore, the average acceleration of the car is $0.834m/{s^2}$.
The correct option is (D), $0.8m/{s^2}$.
Note: We need to take care that the velocity is changed from $km/hr$ to ${{m/s}}$ because the time is given in seconds and the S.I unit of acceleration is ${{m/}}{{{s}}^2}$. The question asks the value of acceleration that is most near to the given options, therefore we rounded off the value of the calculated average acceleration to one-decimal place.
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