Solving a Cart Distance Problem: 18s to 21s

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In summary: Since the initial velocity is 22m/s and the final velocity is 10m/s, the average velocity is (22+10)/2=16m/s. So again, you're not going to get a negative sign from the \frac{1}{2}at^2 formula if you arrange it properly.In summary, the distance traveled by the cart from 18.0s to 21.0s can be calculated by finding the area under the v-t graph. This can be done by breaking it into a triangle and rectangle and adding their respective areas. Another method is using the equation Δx=(v2f−v2i)/2a, where the final and initial velocities are 10m/s and
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bamber296
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Homework Statement


Evaluate the distance the cart travels from 18.0s to 21.0s.

The points are roughly (18.0, 22) and (21.0, 10). The slope/acceleration is constant.


Homework Equations


Δx=(v2f−v2i)/2a
Δx=0.5at2
Δx=0.5(base)(height)


The Attempt at a Solution


I used the first equation as:
Δx=(v2f−v2i)[itex]\overline{}2a[/itex]
=(222−102)/2(-4) = -48 m

Δx=0.5at2 = 0.5(-4)(9) = -18 m

Both of those were wrong along with the last equation (which also equaled -18 m). I really need help!
 
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  • #2


bamber296 said:

Homework Statement


Evaluate the distance the cart travels from 18.0s to 21.0s.

The points are roughly (18.0, 22) and (21.0, 10). The slope/acceleration is constant.

Homework Equations


Δx=(v2f−v2i)/2a
Δx=0.5at2
Δx=0.5(base)(height)

The Attempt at a Solution


I used the first equation as:
Δx=(v2f−v2i)[itex]\overline{}2a[/itex]
=(222−102)/2(-4) = -48 m

Δx=0.5at2 = 0.5(-4)(9) = -18 m

Both of those were wrong along with the last equation (which also equaled -18 m). I really need help!

Answer should be 48m.

Remember, distance (which is what's being asked for) is just an unsigned number. If they asked for displacement, then sign (or direction) is important.

Think of it as area under the v-t graph. It's a trapezoidal area, so it's not just base*height. The area under the graph can be calculated as the sum of a triangle (area: 18) and a rectangle (area:30), though.

What you worked out with [itex]v_f^2 = v_i^2 + 2a{\Delta}x[/itex] is essentially correct, except the final velocity here is 10m/s and the initial velocity is 22m/s. So you wouldn't have got a negative sign if you'd arranged it properly.

The [itex]\frac{1}{2}at^2[/itex] only applies if the object is starting from rest.
 

1. How can I calculate the distance traveled by a cart from 18s to 21s?

The distance traveled by a cart can be calculated using the formula d = v*t, where d is the distance traveled, v is the velocity, and t is the time. In this case, the time is 3 seconds (21s - 18s), so the distance traveled can be found by multiplying the velocity by 3.

2. What is the average velocity of the cart during this time interval?

The average velocity of the cart can be calculated by dividing the total distance traveled (calculated in the previous question) by the total time interval. In this case, the average velocity would be the distance traveled divided by 3 seconds.

3. How can I find the initial velocity of the cart?

To find the initial velocity of the cart, you will need to know the final velocity at 21s, as well as the acceleration of the cart. With this information, you can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval (3 seconds in this case). Rearrange the equation to solve for u, the initial velocity.

4. Is there a specific unit for distance that should be used when solving this problem?

The unit for distance can vary depending on the system of measurement being used. In the International System of Units (SI), the standard unit for distance is meters (m). However, other commonly used units include feet (ft) and kilometers (km). It is important to be consistent with the unit used for distance throughout the problem.

5. Can this problem be solved using kinematic equations?

Yes, this problem can be solved using kinematic equations which are equations that describe the motion of objects. The most commonly used kinematic equations are for constant acceleration, and they include equations for distance, velocity, and acceleration. These equations can be used to solve problems involving motion with constant acceleration, such as the motion of a cart in this problem.

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