Kinematic Equations for Constant Speed

[stevenwho]

Kinematic Equations... Constant Speed

Homework Statement

Question 1) A car pulls away from an intersection when the light turns green. After uniformly accelerating for the next 4.0s, the car has traveled a distance of 50m. The car then proceeds at constant speed.
1. What was the car's acceleration
2. How fast was the car traveleing when it finished accelerating
3. How long will it take for the car to travel another 50 m at this constant speed?

Homework Equations

The 5 kinematic equations

The Attempt at a Solution

For #1... I used a kinematic equation and I found this info
Initial Velocity = 0
Time = 45s
X = 50
a = ?

X = distance
a = acceleration

So then I used x=intitial velocity + acceleration*time
and got 1.11m/s... Am I right?

For #2... I found
Initial Velocity = 0
Time = 4 seconds
X = 50 minutes
a = 1.11m/s
V= ?

Would objects position at time = 0 be relevant here?

I got a answer of 14.72 m/s... Is that right?

For #3
I said X= 100 (50 m + 50 m)
said initial velocity = 0
a = 1.11m/s
t = ?
I got t = 90s... Kind of makes sense, 14.72m/s + 90s = 104 something M...

? - the unknown. Please help :D

Thanks

 

[chocokat]

I think you have a problem right away.

Homework Statement

Question 1) A car pulls away from an intersection when the light turns green. After uniformly accelerating for the next 4.0s, the car has traveled a distance of 50m. The car then proceeds at constant speed.
1. What was the car's acceleration

For #1... I used a kinematic equation and I found this info
Initial Velocity = 0
Time = 45s
X = 50
a = ?

Why are you using 45s as the time? It appears to be 4.0s in your problem. Try recalculating your acceleration.


I'm not sure what this is, and I think you're mixing your formulas up.

2. How fast was the car traveleing when it finished accelerating
3. How long will it take for the car to travel another 50 m at this constant speed?

Homework Equations

The 5 kinematic equations

The Attempt at a Solution

X = distance
a = acceleration

So then I used x=intitial velocity + acceleration*time
and got 1.11m/s... Am I right?

For this #2, you're close, except you'll have to recalculate using the new acceleration you should have found above. The easiest formula to use to find the velocity at that point is:

v_2 = v_1 + at

For #2... I found
Initial Velocity = 0
Time = 4 seconds
X = 50 minutes
a = 1.11m/s
V= ?

Would objects position at time = 0 be relevant here?

I got a answer of 14.72 m/s... Is that right?

For #3 below, I think you've really gone astray. It's only asking for the time for the next 50m, and an important piece of information is:

The car then proceeds at constant speed.

Remember, a constant speed means NO acceleration. So you can just use a constant velocity equation, such as:
v = \frac{dist}{time}

So just toss this stuff below out and try again.

For #3
I said X= 100 (50 m + 50 m)
said initial velocity = 0
a = 1.11m/s
t = ?
I got t = 90s... Kind of makes sense, 14.72m/s + 90s = 104 something M...

? - the unknown. Please help :D

Thanks

Good luck!

 

[stevenwho]

Still can't find the answer to the cars acceleration

 

[chocokat]

Still can't find the answer to the cars acceleration

Well, you need to use the correct equation, which I mentioned above. You have

X = initial velocity + acceleration * time

but that's not the correct equation. The correct equation is:

\Delta x = ut + \frac{1}{2} a t^2

Where u = initial velocity. You can solve for aceleration using
\Delta x = 50m
u = 0
t = 4.0s

So now we have:

50 = (0)(4) + \frac{1}{2} a 4^2

You can do the rest.