Finding velocity, time, and distance

[oliampian]

Homework Statement

A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 17.0 m/s when it reaches the end of the ramp, which has length 117 m .

1) What is the acceleration of the car?
2) How much time does it take the car to travel the length of the ramp?
3) The traffic on the freeway is moving at a constant speed of 17.0 m/s . What distance does the traffic travel while the car is moving the length of the ramp?

Known:
-Vo = 0 (since the car starts from rest)
-Speed = 17.0 m/s
-Distance = 117m

Homework Equations

speed = distance traveled / time
v(t) = v_ot + at
x(t) = x_o + v_ot + 1/2at²

The Attempt at a Solution

1) My attempt at finding the acceleration of the car was figuring out the speed is distance traveled divided by time. So in order to find the time it took for the car to reach the end of the ramp I did:

speed = distance traveled / time
17 = 117 / time
17 * time = 117
time = 117 / 17 = 6.8826

Then I plugged the time I found into the equation v(t) = v_ot + at²:

v(6.8826) = 0(6.8826) + a(6.8826)
17 = a(6.8826)
2.47 = a

2) What I found in question 1:
speed = distance traveled / time
17 = 117 / time
17 * time = 117
time = 117 / 17 = 6.8826

3) I just don't know how to start it. I do think I need to use the position equation.

Both my answers for number 1 and 2 are wrong and I think I'm just not using the right formulas or thinking about this equation correctly. Thanks in advance for any help that I may receive!

 

[Samy_A]

Homework Statement

A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 17.0 m/s when it reaches the end of the ramp, which has length 117 m .

1) What is the acceleration of the car?
2) How much time does it take the car to travel the length of the ramp?
3) The traffic on the freeway is moving at a constant speed of 17.0 m/s . What distance does the traffic travel while the car is moving the length of the ramp?

Known:
-Vo = 0 (since the car starts from rest)
-Speed = 17.0 m/s
-Distance = 117m

Homework Equations

speed = distance traveled / time
v(t) = v_ot + at²
x(t) = x_o + v_ot + 1/2at²

The second "relevant equation" (the one for v(t)) is not correct.

The Attempt at a Solution

1) My attempt at finding the acceleration of the car was figuring out the speed is distance traveled divided by time. So in order to find the time it took for the car to reach the end of the ramp I did:

speed = distance traveled / time
17 = 117 / time
17 * time = 117
time = 117 / 17 = 6.8826

Here you assume that the speed was 17 m/s all the time. That is not correct. 17 m/s is the speed at the end of the ramp.

Look up the correct equation for v(t). That's the equation where you can plug in the 17 m/s.
This, together with the equation for x(t) will give you a and the time for the car to arrive at the end of the ramp.

Then I plugged the time I found into the equation v(t) = v_ot + at²:

v(6.8826) = 0(6.8826) + a(6.8826)²
17 = a(6.8826)²
17 = a(47.3661)
0.3589 = a

2) What I found in question 1:
speed = distance traveled / time
17 = 117 / time
17 * time = 117
time = 117 / 17 = 6.8826

3) I just don't know how to start it. I do think I need to use the position equation.

Both my answers for number 1 and 2 are wrong and I think I'm just not using the right formulas or thinking about this equation correctly. Thanks in advance for any help that I may receive!

 

[SteamKing]

Homework Statement

A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 17.0 m/s when it reaches the end of the ramp, which has length 117 m .

1) What is the acceleration of the car?
2) How much time does it take the car to travel the length of the ramp?
3) The traffic on the freeway is moving at a constant speed of 17.0 m/s . What distance does the traffic travel while the car is moving the length of the ramp?

Known:
-Vo = 0 (since the car starts from rest)
-Speed = 17.0 m/s
-Distance = 117m

Homework Equations

speed = distance traveled / time
v(t) = v_ot + at²
x(t) = x_o + v_ot + 1/2at²

The Attempt at a Solution

1) My attempt at finding the acceleration of the car was figuring out the speed is distance traveled divided by time. So in order to find the time it took for the car to reach the end of the ramp I did:

speed = distance traveled / time
17 = 117 / time
17 * time = 117
time = 117 / 17 = 6.8826

Please show units for all your calculation results.

Why did you use speed = distance / time? You were told that the car is accelerating while trying to get into traffic.

Then I plugged the time I found into the equation v(t) = v_ot + at²:

v(6.8826) = 0(6.8826) + a(6.8826)²
17 = a(6.8826)²
17 = a(47.3661)
0.3589 = a

2) What I found in question 1:
speed = distance traveled / time
17 = 117 / time
17 * time = 117
time = 117 / 17 = 6.8826

You laid out the values for the variables for the right equation in the OP, then you picked the wrong motion equation to find the time required to accelerate up the ramp.

The equation speed = distance / time does not work when acceleration is involved. Speed (velocity) is constantly changing with constant acceleration.

3) I just don't know how to start it. I do think I need to use the position equation.

There should be no great mystery here. The traffic on the freeway is traveling at a constant velocity. All you need to calculate distance traveled is to find the correct time it took the car on the ramp to accelerate to 17.0 m/s

 

[oliampian]

Please show units for all your calculation results.

Why did you use speed = distance / time? You were told that the car is accelerating while trying to get into traffic.

You laid out the values for the variables for the right equation in the OP, then you picked the wrong motion equation to find the time required to accelerate up the ramp.

The equation speed = distance / time does not work when acceleration is involved. Speed (velocity) is constantly changing with constant acceleration.

There should be no great mystery here. The traffic on the freeway is traveling at a constant velocity. All you need to calculate distance traveled is to find the correct time it took the car on the ramp to accelerate to 17.0 m/s

Oh I get it! The equation I should be using is v² = V_o² + 2a(x-x_o)
Then just plugging in the values given would be:
(17m/s)² = 0 + 2a(117m)
-algebra stuff-
a = 1.24 m/s²

Then to find time I would use
x(t) = x_o + v_ot + 1/2at²
117m = 0 + 0 + 1/2(1.24m/s²)t²
-algebra stuff-
t = 13.8 s

Then for the last part constant velocity means there's no acceleration? So then I would use x(t) = x_o + v_ot + 1/2at² again?
x(13.8) = 0 + 17.0 m/s * (13.8 s) + 0
x(13.8) = 234.6 m ... 235m?

 

[Samy_A]

Oh I get it! The equation I should be using is v² = V_o² + 2a(x-x_o)
Then just plugging in the values given would be:
(17m/s)² = 0 + 2a(117m)
-algebra stuff-
a = 1.24 m/s²

Then to find time I would use
x(t) = x_o + v_ot + 1/2at²
117m = 0 + 0 + 1/2(1.24m/s²)t²
-algebra stuff-
t = 13.8 s

Then for the last part constant velocity means there's no acceleration? So then I would use x(t) = x_o + v_ot + 1/2at² again?
x(13.8) = 0 + 17.0 m/s * (13.8 s) + 0
x(13.8) = 234.6 m ... 235m?

Is correct. Your last answer has an error due to rounding: the exact answer is 234 m.