Question

The driver of a car traveling at a speed of 18 m/s slams on the brakes and comes to a stop in 4s . If we assume that the car's speed changed at a constant rate (constant net force): (a) what was the car's average speed during this 4 s interval? (b) How far did the car go in this 4 s interval?

Short answer

(a) Average speed${\mathrm{v}}_{\mathrm{avg}}=9\hspace{0.33em}\mathrm{m}/\mathrm{s}$

(b) Distance travel${\mathrm{d}}_{\mathrm{t}}=36\mathrm{m}$

Step-by-step solution

Identification of given data

• Initial velocity ${\mathrm{v}}_{\mathrm{i}}=18\mathrm{m}/\mathrm{s}$
• Final velocity ${\mathrm{v}}_{\mathrm{f}}=0$
• Time required$\mathrm{t}=4\mathrm{s}$

Average speed

Average speed is ratio of total distance travelled in a particular period of time.

Calculation for the average speed

(a) The average speed of an object is calculated by

$v_{avg}=\frac{d_t}{t}$

Here,

$d_t$ = Total distance

$t$ = Total time

If the speed changes at a constant rate, the average speed is the initial and final velocity arithmetic mean.

$$\begin{gathered} v_{avg}=\frac{v_i+v_f}{2} \\ {v}_{avg}=\frac{18+0}{2} \\ {v}_{avg}=9m/s \end{gathered}$$

Then, The average speed will be$v_{avg}=9m/s$

Calculation for the distance travel

(b) Use the equation refeered to in step 2

$$\begin{gathered} v_{avg}=\frac{d_t}{t} \\ {d}_t=v_{avg}\times t \\ {d}_t=9m/s\times 4s \\ {d}_t=36m \end{gathered}$$

Hence the total distance travel$d_t=36m$