Chapter 3, Problem 6STP

To determine

To Choose: The correct option.

Answer to Problem 6STP

C. 50.0m

Explanation of Solution

Given info:

The given car is initially travelling at $91.0\text{km/h}$ and applied the brakes with an acceleration of $-6.40{\text{m/s}}^{\text{2}}$ .

Restaurant is $\text{30}\text{.0 m}$ ahead.

Formula used:

Final velocity with constant acceleration can be obtained using third equation of motion:

  $v^2{}_f=v^2{}_i+2a\left(d_f-d_i\right)$

Where, $v_f$ is the final velocity, $v_i$ is the initial velocity, a is the acceleration, $d_f$ is the final displacement and $d_i$ is the initial displacement.

Calculation:

Now, it is given that the initial velocity of the car is

  $\begin{array}{c}{v}_i=91.0\text{km/h}\\ \text{=91}\text{.0}\times \frac{1000\text{m}}{3600\text{s}}\\ \text{=25}\text{.27}\text{m/s}\end{array}$

The acceleration acting on the car is

  $a=-6.40{\text{m/s}}^{\text{2}}$

And, finally it stops, thus the final velocity will be

  $v_f=0$

Then by assuming initial distance to be 0m, the stopping distance can be calculated as follows:

  $\begin{array}{l}{v}_f^2=v_i^2+2a\left(d_f-d_i\right)\\ {\left(0\right)}^2={\left(25.27\right)}^2+2\left(-6.40\right)\left(d_f-0\right)\\ d_f=49.89\text{m}\\ d_f\approx 50\text{m}\end{array}$

Conclusion:

Thus, the stopping distance of the car is 50m and correct option is C.