A crate slides across a frictionless horizontal surface with constant acceleration 4.70 m/s2 and over a time interval reaches a final velocity of 13.2 m/s.
(a)
If its initial velocity is 6.60 m/s, what is its displacement (in m) during the time interval? (Indicate the direction with the sign of your answer.)
13.84
 
Correct: Your answer is correct.
 m
(b)
What is the distance it travels (in m) during this interval?
13.84
 
Correct: Your answer is correct.
 m
(c)
If its initial velocity is −6.60 m/s, what is its displacement (in m) during the time interval? (Indicate the direction with the sign of your answer.)
13.81
 
Correct: Your answer is correct.
 m
(d)
What is the total distance it travels (in m) during the interval in part (c)?

Let's go step by step to solve part (d). We need to find the total distance the crate travels during the time interval in part (c), where the initial velocity is \(-6.60 \, \text{m/s}\) and the final velocity is \(13.2 \, \text{m/s}\) with a constant acceleration of \(4.70 \, \text{m/s}^2\).

### Step 1: Identify when the velocity is zero.
The object first decelerates to zero velocity (stopping momentarily) and then accelerates in the positive direction until it reaches \(13.2 \, \text{m/s}\). We can calculate the time it takes for the velocity to reach zero.

Using the equation:
\[
v = v_0 + at
\]
Set \(v = 0\), and solve for \(t\):
\[
0 = -6.60 + (4.70)t
\]
\[
t = \frac{6.60}{4.70} = 1.404 \, \text{seconds}
\]

### Step 2: Find the displacement from \(v_0 = -6.60 \, \text{m/s}\) to \(v = 0\).
Now, we find the displacement during this phase (when the object decelerates to rest):
\[
s_1 = v_0 t + \frac{1}{2} a t^2
\]
\[
s_1 = (-6.60)(1.404) + \frac{1}{2}(4.70)(1.404)^2
\]
Calculating:
\[
s_1 = -9.2664 + 4.6297 = -4.6367 \, \text{m}
\]
So, the displacement until the crate stops is \(-4.64 \, \text{m}\).

### Step 3: Find the displacement from rest to \(v = 13.2 \, \text{m/s}\).
Now, we calculate the displacement after the object starts moving in the positive direction from rest to \(13.2 \, \text{m/s}\). Using the same kinematic equation:
\[
v^2 = v_0^2 + 2as
\]
where \(v_0 = 0\), \(v = 13.2 \, \text{m/s}\), and \(a = 4.70 \, \text{m/s}^2\), we solve for \(s_2\):
\[
(13.2)^2 = 0 + 2(4.70)s_2
\]
\[
s_2 = \frac{(13.2)^2}{2(4.70)} = \frac{174.24}{9.40} = 18.54 \, \text{m}
\]

### Step 4: Find the total distance.
The total distance is the absolute sum of the distances:
\[
\text{Total distance} = |s_1| + s_2 = 4.64 + 18.54 = 23.18 \, \text{m}
\]

### Final Answer:
The total distance traveled during the interval is \(23.18 \, \text{m}\).

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Do you want more details or have any questions?

Here are 5 related questions:
1. How would the displacement change if the acceleration were halved?
2. What is the formula for average velocity during this motion?
3. How would friction impact the calculations in this scenario?
4. How do you calculate the velocity at any given point during the motion?
5. What would happen if the initial velocity and final velocity were both negative?

**Tip:** Always differentiate between *displacement* (a vector quantity) and *distance* (a scalar quantity).

A crate slides across a frictionless horizontal surface with constant acceleration 4.70 m/s² and over a time interval reaches a final velocity of 13.2 m/s. (a) If its initial velocity is 6.60 m/s, what is its displacement (in m) during the time interval? (b) What is the distance it travels during this interval? (c) If its initial velocity is −6.60 m/s, what is its displacement during the time interval? (d) What is the total distance it travels during the interval in part (c)?

Solve a kinematics problem where a crate moves on a frictionless surface with constant acceleration (4.70 m/s²) and initial velocities of 6.60 m/s and -6.60 m/s. Learn to calculate displacement and total distance using the equations of motion. Suitable for high school physics students.

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