Question

With a slope of 6.4%

A car starts at 0mph, and has a top speed of 203
And has a base acceleration of 2.93ft/s2
How long would it take to reach 12.42 miles

Answer 1

Answer: 271.4 s

Step-by-step 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 base acceleration (average acceleration)
`a_(ave)` is:

`a_(ave)=2.93 ft/s^(2)=0.89 m/s^(2)`

Since:

`a_(ave)=(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=(V_(f)-V_(o))/(a_(ave))` (2)

`\Delta t=(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=(V_(f)^(2))/(2a_(ave))` (6)

`d=((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)=(d_(remain))/(t_(2))` (11)

Isolating
`t_(2)`:

`t_(2)=(d_(remain))/(V_(max))` (12)

`t_(2)=(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`