Question

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from top to bottom, how quickly could a car reach the top?
-Slope of 6.4%
-car has top speed 203 mph
A stock car has a peak acceleration of over 1g. Assume this car will accelerate at an average of .091 g (2.93 ft/s2)
The 12.42 mile long Pike’s Peak course has over 156 turns

Answer 1

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

$V_{max}=203 mph=90.74 m/s$ taking into account $1 mile=1609.34 m$

And the car's average acceleration $a_{ave}$ is:

$a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}$

Since:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}$ (1)

Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$:

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$, with the following equation:

$V_{max}=\frac{d_{remain}}{t_{2}}$ (11)

Isolating $t_{2}$:

$t_{2}=\frac{d_{remain}}{V_{max}}$ (12)

$t_{2}=\frac{15362.35 m}{90.74 m/s}$ (13)

$t_{2}=169.45 s$ (14)

With this time $t_{2}$ and the value of $t_{1}$ calculated in (4) we can finally calculate the total time $t_{TOTAL}$:

$t_{TOTAL}=t_{1}+ t_{2}$ (15)

$t_{TOTAL}=101.95 s + 169.45 s$ (16)

$t_{TOTAL}=271.4 s s$