Answer :
Answer: 271.4 s
Explanation:
We are told the top speed (maximum speed) [tex]V_{max}[/tex] the car has is:
[tex]V_{max}=203 mph=90.74 m/s[/tex] taking into account [tex]1 mile=1609.34 m[/tex]
And the car's average acceleration [tex]a_{ave}[/tex] is:
[tex]a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}[/tex]
Since:
[tex]a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}[/tex] (1)
Where:
[tex]V_{f}=V_{max}=90.74 m/s[/tex] is the car's final speed (top speed)
[tex]V_{o}=0 m/s[/tex] because it starts from rest
[tex]\Delta t[/tex] is the time it takes to reach the top speed
Finding this time:
[tex]\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}[/tex] (2)
[tex]\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}[/tex] (3)
[tex]\Delta t=t_{1}=101.95 s[/tex] (4)
Now we have to find the distance [tex]d[/tex] the car traveled at this maximum speed with the following equation:
[tex]V_{f}^{2}=V_{o}^{2} + 2a_{ave} d[/tex] (5)
Isolating [tex]d[/tex]:
[tex]d=\frac{V_{f}^{2}}{2a_{ave}}[/tex] (6)
[tex]d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}[/tex] (7)
[tex]d=4625.70 m[/tex] (8)
On the other hand, we know the total distance [tex]D[/tex] traveled by the car is:
[tex]D=12.42 miles = 19988.052 m[/tex]
Hence the remaining distance is:
[tex]d_{remain}=D-d=19988.052 m - 4625.70 m[/tex] (9)
[tex]d_{remain}=15362.35 m[/tex] (10)
So, we can calculate the time [tex]t_{2}[/tex] it took to this car to travel this remaining distance [tex]d_{remain}[/tex] at its top speed [tex]V_{max}[/tex], with the following equation:
[tex]V_{max}=\frac{d_{remain}}{t_{2}}[/tex] (11)
Isolating [tex]t_{2}[/tex]:
[tex]t_{2}=\frac{d_{remain}}{V_{max}}[/tex] (12)
[tex]t_{2}=\frac{15362.35 m}{90.74 m/s}[/tex] (13)
[tex]t_{2}=169.45 s[/tex] (14)
With this time [tex]t_{2}[/tex] and the value of [tex]t_{1}[/tex] calculated in (4) we can finally calculate the total time [tex]t_{TOTAL}[/tex]:
[tex]t_{TOTAL}=t_{1}+ t_{2}[/tex] (15)
[tex]t_{TOTAL}=101.95 s + 169.45 s[/tex] (16)
[tex]t_{TOTAL}=271.4 s s[/tex]