- #1

- 144

- 30

- Homework Statement
- calculate the distance traveled by a body given the graph

- Relevant Equations
- ##d=V_0\times t + \frac{1}{2}\times at^{2}##, ##d=\frac{vt}{2}##

So basically I wonder why the distance traveled by a body in the 5th second gives different results when calculated by the formula for accelerating body(##d=V_0\times t + \frac{1}{2}\times at^{2}##) and when calculated using a graph(formula for the surface of the triangle).

Here is the graph of the problem(pic below).

When I calculate distance traveled using ##d=\frac{vt}{2}## I get ##d=\frac{\Delta v\Delta t}{2}=\frac{4\times 1}{2}=2m##, and when I do the same using ##d=V_0\times t + \frac{1}{2}\times at^{2}## I get ##d=V_0\times t+\frac{1}{2}\times at^{2} = 2 + (-2)=0## because ##a=\frac{\Delta v}{\Delta t}= -4##.

If someone can please tell me where I am wrong, thank you.

Here is the graph of the problem(pic below).

When I calculate distance traveled using ##d=\frac{vt}{2}## I get ##d=\frac{\Delta v\Delta t}{2}=\frac{4\times 1}{2}=2m##, and when I do the same using ##d=V_0\times t + \frac{1}{2}\times at^{2}## I get ##d=V_0\times t+\frac{1}{2}\times at^{2} = 2 + (-2)=0## because ##a=\frac{\Delta v}{\Delta t}= -4##.

If someone can please tell me where I am wrong, thank you.