Chapter 2: Q26P (page 84)

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?

Expert verified

(a) Average speed $v_{avg} = 9 m / s$

(b) Distance travel $d_{t} = 36 m$

Step by step solution

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

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

$v_{a v g} = \frac{d_{t}}{t}$

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.

$v_{a v g} = \frac{v_{i} + v_{f}}{2} v_{a v g} = \frac{18 + 0}{2} v_{a v g} = 9 m / s$

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

(b) Use the equation refeered to in step 2

$v_{a v g} = \frac{d_{t}}{t} d_{t} = v_{a v g} \times t d_{t} = 9 m / s \times 4 s d_{t} = 36 m$

Hence the total distance travel $d_{t} = 36 m$

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