Acceleration straight line graph

[xzibition8612]

Homework Statement

A particle starts from rest and accelerates as shown in the attached figure. Determine (a) the article's speed at t=10s and t=20s. (b) The distance traveled in the first 20s.

Homework Equations

a=dv/dt

The Attempt at a Solution

(a) t=10s, speed = 20m/s
t=20s, speed = 5m/s
Those were obtained by taking the area under the acceleration.

(b) book gives the answer 262m. NO idea how this came about. I thought about 225m being the answer since 20m/s * 10s + 5m/s * 5s, but this wouldn't be correct because the velocity is changing, not constant. Any help would be appreciated.

 

[ap123]

You can use the constant-acceleration equations to work this out.

 

[xzibition8612]

k is there a way by the graph or integration? I want to use something that's always going to work, if that's possible.

 

[SteamKing]

Graphically, the change in velocity from t1 to t2 represents the area under the acceleration curve.

After constructing a velocity curve, the change in displacement is the area under the velocity curve.

If the acceleration curve is positive, how does that affect velocity?
If the acceleration curve is negative, how does that affect velocity?
If the acceleration curve is zero, how does that affect velocity?

 

[Nickie]

Your solution for part (a) is correct. The speed at t=10s is 20m/s and at t=20s is 5m/s, as determined by taking the area under the acceleration curve. For part (b), the distance traveled can be determined by finding the area under the velocity curve, since distance is equal to velocity multiplied by time. In this case, the distance traveled in the first 20s can be calculated by finding the area of the triangle formed by the velocity points at t=10s, t=20s, and the x-axis. This gives a distance of 225m, as you correctly calculated. It is important to note that the velocity is not constant, but it is still possible to calculate the distance by finding the area under the curve. The answer given in the book may be incorrect or may have been calculated using a different method.