Question

1. a car starts from rest and accelerates to a speed of 60 mph in 20 seconds. the acceleration during this period is constant. for the next 20 minutes, the car moves with the constant speed of 60 mph. at this time the driver of the car applies the brake, and the car decelerates to a full stop in 10 seconds. the variation of the speed of the car with time is shown in the accompanying diagram. determine the total distance traveled by the car and the average speed of the car over this distance. calculate the accelerations of the car in this trip.

Explanation:

Step1: Convert units

First, convert 60 mph to ft/s. Since 1 mile = 5280 ft and 1 hour = 3600 s, then \(v = 60\times\frac{5280}{3600}= 88\) ft/s.

Step2: Calculate acceleration during the first - stage

The formula for acceleration is \(a=\frac{\Delta v}{\Delta t}\). In the first stage, \(v_0 = 0\), \(v = 88\) ft/s, \(t = 20\) s. So \(a_1=\frac{88 - 0}{20}=4.4\) ft/s².

Step3: Calculate the distance during the first - stage

Use the formula \(s_1=v_0t+\frac{1}{2}a_1t^{2}\). Substituting \(v_0 = 0\), \(a_1 = 4.4\) ft/s², \(t = 20\) s, we get \(s_1=\frac{1}{2}\times4.4\times20^{2}=880\) ft.

Step4: Calculate the distance during the second - stage

The time of the second stage \(t_2=20\) minutes \(=20\times60 = 1200\) s, and the speed \(v = 88\) ft/s. Using the formula \(s_2=v\times t_2\), we have \(s_2=88\times1200 = 105600\) ft.

Step5: Calculate acceleration during the third - stage

In the third stage, \(v_0 = 88\) ft/s, \(v = 0\), \(t = 10\) s. Using \(a=\frac{\Delta v}{\Delta t}\), \(a_3=\frac{0 - 88}{10}=- 8.8\) ft/s².

Step6: Calculate the distance during the third - stage

Use the formula \(s_3=v_0t+\frac{1}{2}a_3t^{2}\). Substituting \(v_0 = 88\) ft/s, \(a_3=-8.8\) ft/s², \(t = 10\) s, we get \(s_3=88\times10+\frac{1}{2}\times(-8.8)\times10^{2}=880 - 440=440\) ft.

Step7: Calculate the total distance

\(s=s_1 + s_2+s_3=880+105600 + 440=106920\) ft. Convert to miles: \(s=\frac{106920}{5280}=20.25\) miles.

Step8: Calculate the total time

\(t_{total}=20 + 1200+10=1230\) s. Convert to hours: \(t_{total}=\frac{1230}{3600}\) h.

Step9: Calculate the average speed

The formula for average speed is \(v_{avg}=\frac{s}{t}\). Substituting \(s = 20.25\) miles and \(t=\frac{1230}{3600}\) h, \(v_{avg}=\frac{20.25}{\frac{1230}{3600}}\approx59.2\) mph.

Answer:

• The acceleration in the first stage is \(4.4\) ft/s², the acceleration in the third stage is \(-8.8\) ft/s².
• The total distance traveled by the car is \(20.25\) miles.
• The average speed of the car is approximately \(59.2\) mph.