91Ó°ÊÓ

The relation between the velocity and the acceleration of any object is given with the help of a double derivative, and the acceleration is defined as the rate of change of velocity of an object in unit time.

Given data.

The initial velocity is $v_{\mathrm{i}} = 14 \frac{\mathrm{m}}{\mathrm{s}}$.

The final velocity is $v_{\mathrm{f}} = 21 \frac{\mathrm{m}}{\mathrm{s}}$.

The time taken by the car is $\mathrm{Δ}t = 6.0 \mathrm{s}$.

Assumption.

Let $a$ be the acceleration.

Now, you know that

role="math" localid="1642832676518" $a = \frac{v_{\mathrm{f}} - v_{\mathrm{i}}}{\mathrm{Δ}t}…\left(i\right)$.

According to the second equation of motion,

role="math" localid="1642832716259" ${v_{\mathrm{f}}}^2 = v_{\mathrm{i}}^2 + 2a \mathrm{Δ}x…\left(ii\right)$.

Now, from equation (i),

$\begin{array}{c}a = \frac{v_{\mathrm{f}} - v_{\mathrm{i}}}{\mathrm{Δ}t}\\ = \frac{21 \frac{\mathrm{m}}{\mathrm{s}} - 14 \frac{\mathrm{m}}{\mathrm{s}}}{6.0 \mathrm{s}}\\ = 1.17 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the acceleration of the car is $1.17 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Now, from equation (ii),

$\begin{array}{c}{v_{\mathrm{f}}}^2 = v_{\mathrm{i}}^2 + 2a \mathrm{Δ}x\\  {\left(21 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 = {\left(14 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 + \left(2 × \left(1.17 \frac{\mathrm{m}}{{\mathrm{s}}^2}\right) × \mathrm{Δ}x\right)\\ 2 × \left(1.17 \frac{\mathrm{m}}{{\mathrm{s}}^2}\right) ×\mathrm{Δ}x = {\left(21 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 - {\left(14 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ \mathrm{Δ}x = 104.7 \mathrm{m}\end{array}$

Therefore, the car travels 104.7 m at this time.