Accelerating speed to cover specific distance

[trees1210]

I'm trying to get an equation to calculate how fast an automobile would be going to cover 500 feet if it takes the vehicle 8.9 seconds to go from 0 - 60 mph.

 

[Tac-Tics]

This is a straightforward motion problem.

Do you have any calculus experience? If so, try to pose the problem in terms of the particle's position, its derivative (velocity), and its second derivative (acceleration).

If you don't, we'll get you to that point =-)

 

[trees1210]

It's been way too many years since I took calculus. Any help would be appreciated.

 

[HallsofIvy]

If the constant acceleration is a, then
\frac{dv}{dt}= a
so v= at+ C where C is the initial speed. Since in this problem, v(0)= 0, v(t)= at.

Then we have
\frac{dx}{dt}= at
Since "it takes the vehicle 8.9 seconds to go fr=om 0 - 60 mph", v(8.9)= a(8.9)= 60 so a= 60/8.9= 6.74 "miles per hour per second".
There are 60 minutes per hour and 60 seconds per hour so that is 60(60)= 3600 seconds per hour and so a= 6.74/3600= 0.0019 miles per second per second. But there are 5280 feet per mile so that is 0.0019*5280= 9.98 feet per second per second.
dv/dt= 9.98 so v= 9.98 t when t is measured in seconds and v is measured in feet per second.

Integrating both sides] of v= dx/dt= 9.98 t,
x= (1/2)9.97t^2+ C=.9.85t^2+ C[/itex] feet<br /> and taking the initial positon to be x= 0, C= 0. Now we have x(t)= 9.98t^2<br /> The time take to go 500 feet is given by 9.98t^2= 500, t^2= (500/9.98)= 50.58 and so t= \sqrt{50.58}= 7.11 seconds.

 

[Bob S]

For accelerating automobiles, the acceleration force (wheel torque) is not constant when the transmission is shifted, but the engine power remains constant. As the car speeds up, the wheel RPM increases and the torque decreases. Under this assumption, using W as the constant power, the total energy at time t is

Wt = (1/2)mv².

Using m= 1500 Kg and a speed of 60 mph (26.82 m/s) at 8.9 s, we get

W = (1/2) 1500 26.82²/8.9 = 60,616 watts (= 81.25 HP)

the velocity as a function of time is then

v =dx/dt = sqrt(2Wt/m)

so x(t) by integration is

x(t) = (2/3) sqrt(2W/m) t^3/2

inverting to get t(x) we get

t(x) = [3 x sqrt(m/(8W))]^2/3

so for x = 500 ft (152.4 m), the elapsed time is t =8.64 s.

 

[morrobay]

This problem can be solved with just the basic equations for motion in one dimension:
x=Vot + 1/2at^2
V=Vo+sqrt(2ax)
x=500 ft
60 miles/hr = 88ft/sec so acceleration = 88ft/sec/8.9sec.
a=9.88ft/sec/sec
so 500ft = 9.88(t^2)/2

t=10.06 sec. the velocity at finish line = 99feet/sec
V=67 miles/hour

 

[Bob S]

Both HallsofIvy and morrobay have proposed that the acceleration (i.e. the force) is constant. This leads to a problem.
Given a force F = ma, then work is F dx, and power is F dx/dt = mva

Then, if the acceleration a is constant, then the power is proportional to velocity.

Does this mean that the horsepower of the accelerating car in the OP is six times as much at 60 mph as it was at 10 mph?

 

[morrobay]

Both HallsofIvy and morrobay have proposed that the acceleration (i.e. the force) is constant. This leads to a problem.
Given a force F = ma, then work is F dx, and power is F dx/dt = mva

Then, if the acceleration a is constant, then the power is proportional to velocity.

Does this mean that the horsepower of the accelerating car in the OP is six times as much at 60 mph as it was at 10 mph?

With constant acceleration the power developed is equal to the rate of change of kinetic energy , which of course increases with velocity.

That is, work=Fdx=mxa=1/2mv^2
power=Fdx/dt=mva=1/2mv^2/dt

 

[JCOX]

What about Drag... the square of velocity is proportional to force. So if this is a real life problem it will take a ridiculous more power to accelerate from say 60-100.

 

[Bob S]

What about Drag... the square of velocity is proportional to force. So if this is a real life problem it will take a ridiculous more power to accelerate from say 60-100.

Correct.
The attached thumbnail shows the calculated maximum acceleration speed vs. time for a typical U.S. passenger car. This plot includes air drag and tire rolling resistance, the two biggest power losses (excepting about 10%-15% typ in automatic transmission). The air drag force is proportional to velocity squared, while the tire rolling resistance force is a constant. Details include:
M=1500 Kg
Frontal area = 3 m²
Air density = 1.2 Kg/m³
Air drag coefficient Cp = 0.32
Tire rolling resistance coefficient = 0.01
Horsepower (traction at axle) = 100 HP, (watts = 74,600)

Net power (HP) = traction power – air drag power – tire rolling resistance power

The acceleration rate = net power /(M•v)

At 30 mph, air drag power = 1,388 watts, tire rolling resistance power = 1,972 watts, net HP = 71,240 watts (95.5 HP), acceleration = 3.54 m/s², elapsed time = 1.94 s.

At 60 mph; air drag power = 11,107 watts, tire rolling resistance power = 3,946 watts , net HP = 59550 watts (79.8 HP), acceleration = 1.48 m/s², elapsed time = 8.10 s.

At 90 mph; air drag power = 37,490 watts, tire rolling resistance power = 5,920 watts, net HP (watts) = 31,190 watts (41.8 HP), acceleration = 0.52 m/s², elapsed time = 23.20 s.

Note that at 90 mph, over half the traction HP is being lost in air drag and tire rolling resistance.