A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.
What is the acceleration of the sled?
What is the speed of the sled when it passes the 14.4-m point?
How much time did it take to go from the top to the 14.4-m point?
How far did the sled go during the first second after passing the 14.4-m point?
Acceleration = .800 m/(s^2)
Speed passed 14.4 = 4.8 m/s
Time at 14.4 = 6 s
Distance = 5.20 m
To find the acceleration of the sled, we can use the second equation of motion:
\[
d = ut + \frac{1}{2}at^2
\]
where:
- \(d\) is the distance traveled,
- \(u\) is the initial velocity (which is 0, as the sled starts from rest),
- \(a\) is the acceleration, and
- \(t\) is the time.
Let's use the given information to find the acceleration of the sled:
From the problem, we know:
- At \(t = 0\), the distance traveled (\(d\)) is 0.
- At \(t = 2.00\) s, the distance traveled (\(d\)) is 14.4 m.
- At \(t = 4.00\) s, the distance traveled (\(d\)) is 25.6 m.
- At \(t = 6.00\) s, the distance traveled (\(d\)) is 40.0 m.
- At \(t = 8.00\) s, the distance traveled (\(d\)) is 57.6 m.
We can plug in these values into the equation and solve for the acceleration (\(a\)).
First, let's find the acceleration:
For the time interval from \(t = 0\) s to \(t = 2.00\) s:
\[
14.4 = 0 \times 2.00 + \frac{1}{2}a(2.00)^2
\]
Simplifying:
\[
14.4 = 2.00a
\]
So, \(a = \frac{14.4}{2.00}\).
\(\therefore\) the acceleration of the sled is \(\frac{14.4}{2.00}\) m/s².
To find the speed of the sled when it passes the 14.4 m point, we can use the first equation of motion:
\[
v = u + at
\]
where:
- \(v\) is the final velocity,
- \(u\) is the initial velocity (which is 0, as the sled starts from rest),
- \(a\) is the acceleration, and
- \(t\) is the time.
Let's use the given information to find the speed of the sled:
At \(t = 2.00\) s, we need to find the final velocity (\(v\)).
Using the equation:
\[
v = 0 + a \times 2.00
\]
Substituting the value of acceleration that we found, we get:
\[
v = \frac{14.4}{2.00} \times 2.00
\]
Simplifying:
\[
v = 14.4\, \text{m/s}
\]
\(\therefore\) the speed of the sled when it passes the 14.4 m point is 14.4 m/s.
To find the time it took to go from the top to the 14.4 m point, we can rearrange the second equation of motion:
\[
d = ut + \frac{1}{2}at^2
\]
to solve for time (\(t\)):
\[
2dt + at^2 - 2d = 0
\]
Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can solve for \(t\).
In this case, substituting:
- \(a = \frac{1}{2} \times (\text{acceleration})\),
- \(b = 2 \times (\text{acceleration}) \times \text{distance}\),
- \(c = -2 \times \text{distance}\),
we can find the time it took to go from the top to the 14.4 m point.
Using the given values:
\(a = \frac{1}{2} \times \frac{14.4}{2.00}\),
\(b = 2 \times \frac{14.4}{2.00} \times 14.4\),
\(c = -2 \times 14.4\),
we can solve for \(t\).
\(\therefore\) the time it took to go from the top to the 14.4 m point is the positive value obtained from the quadratic formula.
Lastly, to find out how far the sled went during the first second after passing the 14.4 m point, we can use the first equation of motion:
\[
d = ut + \frac{1}{2}at^2
\]
where:
- \(d\) is the distance traveled,
- \(u\) is the initial velocity (which we can calculate using the final velocity from the previous step),
- \(a\) is the acceleration, and
- \(t\) is the time interval.
Using the equation:
\[
d = u \times 1 + \frac{1}{2}a(1)^2
\]
Substituting the calculated acceleration and the speed at the 14.4 m point, we can solve for the distance (\(d\)).
\(\therefore\) the distance the sled traveled during the first second after passing the 14.4 m point is the value obtained from the equation.
To find the answers to the given questions, we can use kinematic equations that describe the motion of an object with constant acceleration.
Let's start with the first question:
1. What is the acceleration of the sled?
We can use the equation:
d = ut + (1/2)at^2
Here, d is the distance traveled, u is the initial velocity (which is zero since the sled starts from rest), a is the acceleration, and t is the time.
We are given the following distances and time intervals:
d1 = 14.4 m, t1 = 2.00 s
d2 = 25.6 m, t2 = 4.00 s
d3 = 40.0 m, t3 = 6.00 s
d4 = 57.6 m, t4 = 8.00 s
Let's calculate the acceleration using the second and fourth data points:
Using data points 2 and 4:
d4 - d2 = u(t4 - t2) + (1/2)a(t4 - t2)^2
57.6 m - 25.6 m = 0 + (1/2)a(8.00 s - 4.00 s)^2
32.0 m = 4a(4.00 s)^2
32.0 m = 16a(16.00 s^2)
a = (32.0 m) / (16 * 16.00 s^2)
a = 0.50 m/s^2
Therefore, the acceleration of the sled is 0.50 m/s^2.
Moving on to the second question:
2. What is the speed of the sled when it passes the 14.4-m point?
To find the speed, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and t is the time.
We can use the first data point:
v = 0 + (0.50 m/s^2)(2.00 s)
v = 1.00 m/s
Therefore, the speed of the sled when it passes the 14.4-m point is 1.00 m/s.
Next, let's answer the third question:
3. How much time did it take to go from the top to the 14.4-m point?
To find the time, we can rearrange the equation used in the previous question:
v = u + at
t = (v - u) / a
Using the velocity obtained in the previous question:
t = (1.00 m/s - 0) / 0.50 m/s^2
t = 2.00 s
Therefore, it took 2.00 seconds for the sled to go from the top to the 14.4-m point.
Finally, let's answer the fourth question:
4. How far did the sled go during the first second after passing the 14.4-m point?
To find the distance, we can use the equation:
d = ut + (1/2)at^2
We need to find the distance covered during the first second after passing the 14.4-m point, which is from t = 2.00 s to t = 3.00 s.
Using the acceleration and the time interval:
d = 0 m/s * 1.00 s + (1/2)(0.50 m/s^2)(1.00 s)^2
d = 0 m
Therefore, the sled did not cover any distance during the first second after passing the 14.4-m point.