How Do You Calculate Acceleration and Speed in a Joy Ride Physics Problem?

[Mspike6]

Hey,
i got this question in my first assignment in PHYS 109. i am sure the question is easy, but for some reason i can't get it around my head.
Thanks for the help !

Homework Statement

A physics 109 student is taking a car for a joy ride on a long, empty and perfectly
straight Saskatchewan highway. At the beginning of the joy ride the student has the
car accelerate from rest to some top speed with a (constant) acceleration of 4:50 m=s2
and then maintains that top speed to the end of the ride. The total time taken from
the start of the joy ride to its end is 1:36 minutes and the total distance covered is
3:84 km.
(a) What was the time elapsed during the acceleration phase of the car's motion?
(b) What distance did the car cover during the acceleration phase of its motion?
(c) What was the top speed reached by the car?

Homework Equations

V = Vi +at
Xf=Xi +Vi(t) + (1/2)(a)(t^2)
V=(Vf+vi)/2

The Attempt at a Solution

Givens:
a=4.50
t=81.6 s
Xf =3.84 * 10^3 m

a) Vavg. = X/t = 3480/81.6 = 47.1 m/s

Vavg= (Vf + Vi)/2
=> Vf= 2(Vavg - Vi)
==>Vf= 2(47.1 -0)= 94.1 m/s

Vf=at +Vi
=>t=(Vf - Vi)/a = 20.9 s


b)
Xf=Xi +Vi(t) + (1/2)(a)(t^2)

Xf= 0+0+(0.5)(4.50)(20.9^2) = 982.8m



C)Vf = 94.1 m/s




i don't believe in my answer because :
1) The top speed is 339 KM/h...yes it is possible, but he must be one rich Phys student we are talking about here

2) I got the Vf in the very bigening, yet he asks me about it at the end ? then i must be doing somethiong wrong



Thank you guys, anhy help is really appreciated

 

[songoku]

a) Vavg. = X/t = 3480/81.6 = 47.1 m/s

Hi mspike

You can't use x/t to find the speed because there is acceleration.

Now consider two cases:
1. when the car is accelerating
2. when the car moves with constant speed after accelerating

t₁+t₂ = 81.6 s
s₁+s₂ = 3.84 x 10³m

Find s₁ (distance traveled when accelerating) in term of t₁, then find also s₂ (distance traveled with constant speed) in term of t₁.
Use s₁+s₂ = 3.84 x 10³m, then you'll get quadratic equation in term of t₁. By solving that, you'll get two values for t₁; reject the impossible one

 

[Mspike6]

First of all thank you for your replay.

i understood what you mean, but i don't know how to actually do it (i have been tryiong since you replaied)

i don't know how to "Find s1 (distance traveled when accelerating) in term of t1, then find also s2 (distance traveled with constant speed) in term of t1. "the closed thing i could get to is.

T1 + T2 = 81.6
T1= 81.6 - T2

S1f = S1i + V1i (T1) +(o.5)(a)(T1)^2

S1f = 0 + 0 + (2.25)(81.6 - T2)^2
S1f = (2.25) ( 6658.56 - 163.2 T2 + T2^2)

You see where am going ? but, then the final answers are all impossible.if you could give me 1 or 2 more hints.

Thanks

 

[songoku]

First of all thank you for your replay.

i understood what you mean, but i don't know how to actually do it (i have been tryiong since you replaied)

i don't know how to "Find s1 (distance traveled when accelerating) in term of t1, then find also s2 (distance traveled with constant speed) in term of t1. "


the closed thing i could get to is.

T1 + T2 = 81.6
T1= 81.6 - T2

S1f = S1i + V1i (T1) +(o.5)(a)(T1)^2

S1f = 0 + 0 + (2.25)(81.6 - T2)^2
S1f = (2.25) ( 6658.56 - 163.2 T2 + T2^2)

You see where am going ? but, then the final answers are all impossible.


if you could give me 1 or 2 more hints.

Thanks

Actually you expressed s₁ in term of t₂ (see your last equation )

It also works. Now state s₂ in term of t₂. What is the formula to find s₂?

 

[Mspike6]

I would say the same equation as before

S2f = S2i + V2i (T2) +(1/2)(a)(T2)^2

S2f = S2i + V2i (T2) + 0

V2i = V1f Dead end .

Is there a better formula to use ?

 

[songoku]

I would say the same equation as before

S2f = S2i + V2i (T2) +(1/2)(a)(T2)^2

S2f = S2i + V2i (T2) + 0

V2i = V1f


Dead end .

Is there a better formula to use ?

You are on the right track. Note that S2i = 0. Can you find V1f in term of t₂?

 

[Mspike6]

V1f = V2i

V1f = aT1 +V1i

V1f = 4.50 ( 81.6 - T2) +0
V1f = 367.2 - 4.50 T2

Since V1f = V2i

Therefor, S2f = S2i + (367.2 - 4.50 T2) (T2)

S2f = 0 + 367.2 T2 - 4.50 T2 ^2 *Shouldn't S2i be equal to S1f ? but am takeing your word for it*

S1 + S2 = 3840

(T2² -367.2 T2 + 14981.85 + 367.2 T2 - 4.50T2² = 3480

-4.50T2² + 0 T2 + 14981.85-3840 = 0

T2= 50.6 s

wow..

 

[Mspike6]

After finishing the calculation i think there is still something wrong.

If T2 = 50,6s then T1 = 31s

Than to Calculate S1f we will say

S1f = S1i + V1i + (1/2)(a)(T1^2)

S1f = 2162.25 m (which makes no sense.. but still let's carry on)

to Find Top speed (V1f)

V1f² = V1i² + 2 (Delta x)a

Which is equal to 139.5 m/s these guys must have one heck of a car ! :D

 

[songoku]

S2f = 0 + 367.2 T2 - 4.50 T2 ^2 *Shouldn't S2i be equal to S1f ? but am takeing your word for it*

No, S2i does not equal to S1f. We have broken the problem into two cases. So, the initial distance for each case is zero.

S1 + S2 = 3840

(T2² -367.2 T2 + 14981.85 + 367.2 T2 - 4.50T2² = 3480

Check the s₁ again

 

[Mspike6]

Thank you so much for your help Songoku !