Question

A racing car starts from rest at $\mathit{t}\mathbf{=}\mathbf{0}$ and reaches a final speed $\mathit{v}$at time t . If the acceleration of the car is constant during this time, which of the following statements are true?

(a) The car travels a distance $\mathit{v}\mathit{t}$ .

(b) The average speed of the car is $\mathit{v}\mathbf{/}\mathbf{2}$.

(c) The magnitude of the acceleration of the car is$\mathit{v}\mathbf{/}\mathit{t}$ .

(d) The velocity of the car remains constant.

(e)None of statements (a) through (d) is true.

Short answer

Hence, options (b) and (c) are correct.

Step-by-step solution

Given value in the question

The acceleration of the car in the initial state is, $v_0=0 \text{m/s}$,

Final speed of the car$=v$

The expression used to calculate the acceleration of the car is given below.

role="math" localid="1663735439132" $\mathit{a}\mathbf{=}\frac{\mathbf{v}\mathbf{-}{\mathbf{v}}_{\mathbf{0}}}{\mathbf{\Delta }\mathbf{t}}$

Substitution of the values in the expression

Put the given value in the expression is:

$$\begin{gathered} a=\frac{v-0\text{m}/\text{s}}{t} \\ =\frac{v}{t} \end{gathered}$$

So, option (C) is correct.

The displacement of the car to use the following kinetic equation will be

$v^2=v_0^2+2ax$

The above equation will be rearranged for x ,that is

$$\begin{gathered} x=\frac{v^2-v_0^2}{2a} \\ x=\frac{v^2-{(0 \text{m/s})}^2}{2\left(\frac{v}{t}\right)} \\ =\frac{vt}{2} \end{gathered}$$

Determining the value of average speed

The average speed of the car is:

$$\begin{gathered} v_{avg}=\frac{x}{t} \\ =\frac{vt}{2t} \\ =\frac{v}{2} \end{gathered}$$

So, option (b) is correct.