3. a car starts from rest and accelerates to a speed of 60 mph in 20 seconds. the acceleration during this…
3. 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.
Answer
Explanation:
Step1: Convert units
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
Use the formula (a=\frac{v - u}{t}), where (u = 0) (starts from rest), (v = 88) ft/s, (t = 20) s. So (a_1=\frac{88 - 0}{20}=4.4) ft/s².
Step3: Calculate distance during the first - stage
Use the formula (s_1=ut+\frac{1}{2}a_1t^{2}), with (u = 0), (a_1 = 4.4) ft/s², (t = 20) s. Then (s_1=\frac{1}{2}\times4.4\times20^{2}=880) ft.
Step4: Calculate distance during the second - stage
The time (t_2=20) minutes (=20\times60 = 1200) s, speed (v = 88) ft/s. Using (s_2=v\times t_2), we get (s_2=88\times1200 = 105600) ft.
Step5: Calculate deceleration during the third - stage
Use the formula (a=\frac{v - u}{t}), where (u = 88) ft/s, (v = 0), (t = 10) s. So (a_3=\frac{0 - 88}{10}=- 8.8) ft/s².
Step6: Calculate distance during the third - stage
Use the formula (s_3=ut+\frac{1}{2}a_3t^{2}), with (u = 88) ft/s, (a_3=-8.8) ft/s², (t = 10) s. Then (s_3=88\times10+\frac{1}{2}\times(-8.8)\times10^{2}=880 - 440=440) ft.
Step7: Calculate total distance
(s=s_1 + s_2+s_3=880+105600 + 440=106920) ft.
Step8: Calculate total time
(t=t_1 + t_2+t_3=20 + 1200+10=1230) s.
Step9: Calculate average speed
Use the formula (v_{avg}=\frac{s}{t}), so (v_{avg}=\frac{106920}{1230}=87) ft/s.
Answer:
- Total distance: (106920) ft.
- Average speed: (87) ft/s.
- Acceleration in the first stage: (4.4) ft/s².
- Deceleration in the third stage: (-8.8) ft/s².