Question

A driver starts from rest on a straight test track that has markers every0.1 Km. The driver press on the accelerator for the entire period of the test holds the car at constant acceleration. The car passes the0.1 Kmpost at 8 sec after starting the test. (a) What was the car’s acceleration? (b) What was the cars speed as it passed the 0.1 Km post? (c) What does the speedometer read at the post? (d) When does the car pass the 0.2 Km post?

Short answer

The acceleration of the car is $3m/s^2$.

Step-by-step solution

Step 1: Given data

Distance, s = 0.1 Km

Time, t = 8s

Post distance = 0.1 Km

Step 2: Introduction

Acceleration is defined as the increase in speed or rate of any motion.

Step 3: Find the car’s acceleration

The kinematic equations of motion for a particle can be expressed as follow:

$\begin{array}{l}v=u+\alpha t\\ s=S_0+ut+\frac{1}{2}\alpha t^2\\ v^2-u^2=2\alpha \mathrm{s}\end{array}$

Here, s is the distance travelled, $s_0$is the initial distance of the particle, u is the initial velocity, $\alpha$is the acceleration, v is the final velocity, and t is the time taken.

Calculate the acceleration of the car as follows:

From the given data, the car is at rest initially.

That is, u = 0 and $s_0=0$

And the car covers a distance s = 0.1 Km in time t = 8 s .

Rearrange the equation $S=ut+\frac{1}{2}\alpha t^2$for $\alpha$as follows:

$\begin{array}{l}s=ut+\frac{1}{2}\alpha t^2\\ \frac{1}{2}\alpha t^2=s-ut\\ \alpha =\frac{2(s-ut)}{t^2}\end{array}$

Substitute 0m/s for u , 0.1 km for s , and 8s for t in equation $\alpha =\frac{2\left(S-ut\right)}{t^2}$

$$\begin{gathered} \alpha =\frac{2\left[0.1Km\left({\displaystyle \frac{1000m}{1km}}\right)-\left(0m/s\right)\left(8s\right)\right]}{{\left(8s\right)}^2} \\ \alpha =3.125m/s^2 \\ \alpha =3m/s^2 \end{gathered}$$

Therefore, the acceleration of the car is $3m/s^2$.