What is the Velocity of a Car at the End of the Initial 2.09-Second Interval?

[Physicsnoob90]

Homework Statement

A car is traveling along a straight road at a velocity of +34.2 m/s when its engine cuts out.
For the next 2.09 seconds, the car slows down, and its average acceleration is a1. For the
next 6.49 seconds, the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the 8.58-second period is +17.5 m/s. The ratio of the
average acceleration values is a1/a2 = 1.67. Find the velocity of the car at the end of the initial 2.09-second interval.

Homework Equations

Using the kinematic equations

part 1:
vo = 34.3m/s
t=2.09s
vf=v1
a=a1

part 2:
v0=v1
vf=17.5m/s
t=6.49s
a=a2

The Attempt at a Solution

1. using v=vo + at for part 1, v1 =34.2m/s + (1.67a1)(2.09s)
2. plug in what v1 equal to part 2

 

[happysmiles36]

what is the link between a₁ and a₂? Can you think of a way to replace a₂ once you plug in what v₁ equal to part 2?

 

[Physicsnoob90]

well i initially made a1=1.67(a2) but I'm having a hard time finding out how to plug in V1 into part 2.

so far my equation for part 2 is: 17.5m/s = V1+6.49 s (a2)

 

[happysmiles36]

If you make a₁ = 1.67a₂, then when you replace the a₁, you'd be solving for a₂ and that acceleration is in the 6.49s part not the 2.09s part the question is asking for.

V=v_o +at

in

V₂=v_o + a₂t₂
What equation does the v_o represent?

Edit:
To help you see the connection better:

v₁ =34.2m/s + (a₁)(2.09s)
17.5m/s = V₁+(a₂)(6.49 s )

In other words, Can you connect these two equations into one?

 

[Physicsnoob90]

Since V0=V1, does V1 = 34.2m/s +(1.67a2)(2.09s) = 34.2m/s + 3.4903a2

 

[happysmiles36]

Yes v_o=v₁

So substitute the equation for v₁ into v_o

(use the equations under my edit: because your part 1 equation is wrong.)
(You had acceleration in your part 1 as 1.67a₁ when it just needs a₁)

 

[Physicsnoob90]

I'm sorry if I'm making some algebra mistake but would putting part 1 equation into part 2 be:

17.5m/s=(34.2m/s +2.09s a1) +6.49s a2

 

[happysmiles36]

Thats correct.

You can replace a2 with a1/1.67. Solve for a1 and plug it into the first equation for your speed after the 2.09s mark.

(34.2m/s + 2.09s a1)

 

[Physicsnoob90]

so far i have:

17.5m/s = (34.2m/s +2.09s (1.67a2)) + 6.49s a2

-16.7m/s = 2.09s(1.67a2) + 6.49s a2

-16.7m/s = 3.4903sa2 + 6.49sa2

-16.7m/s = 9.9803sa2
-1.67 = a2

 

[happysmiles36]

You need to replace a2 with (a1/1.67) not the other way around. (i explained this in a earlier post)

17.5 = (34.2 + 2.09(a1)) + 6.49((a1)/1.67)

Now solve for a1 and once you find that use this equation (from part 1) to get your velocity after the 2.09s mark.

(34.2m/s + 2.09s a1)

If you need help with this lmk. :)

 

[Physicsnoob90]

so would a1 = -3.2504662

17.5 = 34.2 + 2.09a1 + 6.49 (a1/1.67)
subtracted 17.5 by 34.2
-16.7 = 2.09a1 + 6.49 (a1/1.67)
multiply both side by 1.67
-27.889 = 2.09a1 + 6.49a1
add both a1 and divide the number by-27.889
a1= -3.2504662

 

[happysmiles36]

-16.7 = 2.09a1 + 6.49 (a1/1.67)
multiply both side by 1.67
-27.889 = 2.09a1 + 6.49a1

Here you made a mistake. If you choose to multiply 1.67, you have to do that to all terms. You missed the 2.09 term. I would just take 6.49/1.67 a1. Pull the a1 out and add the 2.09 and (6.49/1.67) together and divide.

a1 should be something like -2.79

Edit:

-16.7 = a1(2.09+(6.49/1.67))

should look something like that if I didn't explain that clear enough.

 

[Physicsnoob90]

Correct a1= -2.79264...

so from here it would be:

v1= 34.2+2.09(-2.79) = 28.3689

 

[happysmiles36]

yep, that is what I got, don't forget units and significant digits

 

[Physicsnoob90]

Big Thanks. really appreciated