Finding distance given two different velocities and times?

[sam1865]

Homework Statement

On a straight track, a cart travels with a constant velocity of 2.5m/s for 9.0s. It then uniformly accelerates to a final velocity of 6.0m/s in 15s. Calculate the total distance traveled by the cart.

Homework Equations

Vavg = Vi+Vf/2
d=(Vf+Vi/2)x t
d=Vi x t + 1/2at^2
a=Vf-Vi/t

The Attempt at a Solution

I tried finding the average velocity by adding together 6.0 and 2.5 before dividing it by 2 to get 4.25. Then, I multiplied that number by the total time of 24s (15+9). This got me the number 102, but the answer key says the answer is 86?

 

[SteamKing]

Homework Statement

On a straight track, a cart travels with a constant velocity of 2.5m/s for 9.0s. It then uniformly accelerates to a final velocity of 6.0m/s in 15s. Calculate the total distance traveled by the cart.

Homework Equations

Vavg = Vi+Vf/2
d=(Vf+Vi/2)x t
d=Vi x t + 1/2at^2
a=Vf-Vi/t

The Attempt at a Solution

I tried finding the average velocity by adding together 6.0 and 2.5 before dividing it by 2 to get 4.25. Then, I multiplied that number by the total time of 24s (15+9). This got me the number 102, but the answer key says the answer is 86?

IDK why you chose this approach. The problem clearly states that the cart accelerates from 2.5 m/s to a final velocity of 6.0 m/s, which means that during the acceleration, average velocities will not give correct distances. You state the proper formulas in your post; why not use them?

You need to work this problem in two phases: 1. the portion where the velocity is constant, and 2. the portion where the car is accelerating.

 

[DaveC426913]

More succinctly, why did you multiply the average speed of the second (accelerative) leg of the journey by the duration of the entire journey?

 

[Juxtaroberto]

The formulas you are using are for constant acceleration, but there isn't constant acceleration overall. There is zero acceleration for the first 9.0 s, and then there is non-zero acceleration from then until the end of the problem. You cannot apply these formulas over the entire interval in question, but you can apply them to each interval alone. This is one way of doing the problem.

Perhaps a more intuitive and easier-to-see approach would be to graph the velocity as a function of time (you'd have a piece-wise graph) and then integrate your graph over the entire interval (you'd have to do two integrals). This would amount to finding the area under the graph, and the total area under the graph will be the total displacement.

 

[HalfLostAndSearching]

I would first clarify the problem and make sure I understand the given information correctly. From the given information, it seems that the cart initially travels with a constant velocity of 2.5m/s for 9.0s, then accelerates uniformly to a final velocity of 6.0m/s in 15s. It is not clear if the cart maintains this final velocity for any amount of time after the 15s period.

Assuming that the cart maintains a final velocity of 6.0m/s after the 15s period, we can calculate the distance traveled in two parts: the distance traveled during the initial 9.0s with a constant velocity of 2.5m/s, and the distance traveled during the acceleration phase of 15s.

For the first part, we can use the equation d = vt, where v is the constant velocity of 2.5m/s and t is the time of 9.0s. This gives us a distance of 22.5m.

For the second part, we can use the equation d = vit + 1/2at^2, where vi is the initial velocity of 2.5m/s, a is the acceleration, and t is the time of 15s. To find the acceleration, we can use the equation a = (vf - vi)/t, where vf is the final velocity of 6.0m/s and t is the time of 15s. This gives us an acceleration of 0.17m/s^2. Plugging this into the first equation, we get d = 2.5(15) + 1/2(0.17)(15)^2 = 37.5 + 19.125 = 56.625m.

Adding the two distances together, we get a total distance of 22.5m + 56.625m = 79.125m, which is closer to the given answer of 86m. This difference could be due to rounding errors or if the cart maintains a final velocity of 6.0m/s for some amount of time after the 15s period. Therefore, it would be best to clarify the problem and make sure all the given information is accurate before solving.