Distance, Velocity, and Acceleration of a vehicle

mcwang719

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The acceleration of a vehicle takes the form du/dt =3.6 - 0.06u where u is the vehicle speed in ft/sec. The vehicle is traveling at 45 ft/sec at time T.

a) Determine the distance traveled by the vehicle when accelerated to a speed of 55 ft/sec.

b) Determine the time at which the vehicle attains the speed of 55 ft/sec.

c) Determine the acceleration of the vehicle after 3 sec.

I solved the problem and just wanted someone to confirm that it's correct. For a) i got x= 81.47 ft and for b) i got t=1.78sec and c) a =0.85 ft/s.

I used the 3 equations that were given in the book:
x=1/2at^2+Vt
u= (alpha/beta)(1-e^(-beta*t)) + u_oe^(-beta*t)
x= (alpha/beta)t - (alpha/beta^2) (1-e^(-beta*t) + (u_o/beta)(1-e^(-beta*t))
du/dt= alpha - beta*u

Can someone please tell me if I did this correctly. Thanks!!!!!!
 
mcwang719 said:
The acceleration of a vehicle takes the form du/dt =3.6 - 0.06u where u is the vehicle speed in ft/sec. The vehicle is traveling at 45 ft/sec at time T ( = 0 ?).

a) Determine the distance traveled by the vehicle when accelerated to a speed of 55 ft/sec.

b) Determine the time at which the vehicle attains the speed of 55 ft/sec.

c) Determine the acceleration of the vehicle after 3 sec.

I solved the problem and just wanted someone to confirm that it's correct. For a) i got x= 81.47 ft and for b) i got t=1.78sec and c) a =0.85 ft/s.

If you show us the steps - we can tell whether you have done it correctly or not.

I used the 3 equations that were given in the book:

x=1/2at^2+Vt..........This is not applicable for this problem

u= (alpha/beta)(1-e^(-beta*t)) + u_oe^(-beta*t)
x= (alpha/beta)t - (alpha/beta^2) (1-e^(-beta*t) + (u_o/beta)(1-e^(-beta*t))
du/dt= alpha - beta*u
The above 3 equations are applicable
Can someone please tell me if I did this correctly. Thanks!!!!!!
 
b) used the first equation. and plugged in the numbers 55= (3.6/.06)(1-e^(-0.6t) +45e^(-.06t) and got t=1.78 seconds

Then used the answer to to get a) plugged t into equation 2. x=(3.6/0.06)1.78-(3.6/0.06^2)(1-e^(-0.06*1.78)) +(45/0.06)(1-e^(-0.06*1.78)) solved for x to get the distance of 81.47 feet.

And for c) to get the acceleration I plugged t=3 seconds into the 2nd equation to get a distance of 138.82 ft. Then plugged it into the equation x=1/2at^2+vt ======> 138.82=1/2a(3)^2+45(3) to get a= 0.85ft/s
 
mcwang719 said:
b) used the first equation. and plugged in the numbers 55= (3.6/.06)(1-e^(-0.06t) +45e^(-.06t) and got t=1.78 seconds

55 = 60 - 15 * e^(-0.06t)

0.06t = ln(3)

t = ln(3)/0.06 = 18.3102

You should check your answer by using graphing calculator




Then used the answer to to get a) plugged t into equation 2. x=(3.6/0.06)1.78-(3.6/0.06^2)(1-e^(-0.06*1.78)) +(45/0.06)(1-e^(-0.06*1.78))
solved for x to get the distance of 81.47 feet.

And for c) to get the acceleration I plugged t=3 seconds into the 2nd equation to get a distance of 138.82 ft. Then plugged it into the equation

x=1/2at^2+vt this equation is ONLY valid for CONSTANT ACCELERATION - this problem is NOT constant acceleration

======> 138.82=1/2a(3)^2+45(3) to get a= 0.85 ft/s ... wrong unit for acceleration
 
sorry i messed up on the math. I did it again and got t=18.31 as well. So are my other steps correct I know now that I have to plug in 18.31 instead of 1.78, so if I make those corrections will the problem be correct? Thanks!!!!!!
 
mcwang719 said:
sorry i messed up on the math. I did it again and got t=18.31 as well. So are my other steps correct I know now that I have to plug in 18.31 instead of 1.78,

so if I make those corrections will the problem be correct? No -

it seems that you have not read my response fully.


Thanks!!!!!!
 
ok. so are a and b correct? I was thinking if I plugged in t=3 into the first equation to get u. then put that value into 3.6-0.06u=a.

So u= (3.6/.06)(1-e^(-.06*3))+45e^(-0.06*3)) to get u=47.47ft/s

Then du/dt=3.6-0.06(47.47) =0.75 ft/s^2.

thanks!!!
 
sorry i messed up on the math. I did it again and got t=18.31 as well. So are my other steps correct I know now that I have to plug in 18.31 instead of 1.78, so if I make those corrections will the problem be correct? Thanks!!!!!

How did t=18.31 become t=1.78? Thank you
 
Please, please read carefully.

The very first thing SK asked is whether T = 0. I’d ask whether you meant that velocity is 45 ft/sec when t = 0. Capital T is an undefined variable that appears nowhere in your equations.

Frankly, I would not bother to work on this question until you answered that initial question.

Later you ask how t = 18.31 become t = 1.78? The 1.78 was YOUR value, and you did not bother to explain how you computed it.
 
Please, please read carefully.

The very first thing SK asked is whether T = 0. I’d ask whether you meant that velocity is 45 ft/sec when t = 0. Capital T is an undefined variable that appears nowhere in your equations.

Frankly, I would not bother to work on this question until you answered that initial question.

Later you ask how t = 18.31 become t = 1.78? The 1.78 was YOUR value, and you did not bother to explain how you computed it.
That is probably because this thread is ~15 years old.
 
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