Question

A particle is moving with a constant acceleration of 4.0 m/s 2. Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s?

Answer 1

We know that the acceleration is defined as:

`a=(dv)/(dt)`

This means that if we know the acceleration we can find the velocity with the integral:

`v=\int ^t_(t_0)adt^(\prime)+v_0`

where t0 denotes some intial time and v0 the velocity at that time.

In this case we know that the acceleration is 4m/s^2 and that at time t=1 the velocity is 4 m/s, then we have:

`\begin{gathered} v=\int ^t_14dt^(\prime)+4 \\ =4(t^(\prime))\vert^t_1+4 \\ =4(t-1)+4 \\ =4t-4+4 \\ =4t \end{gathered}`

hence the velocity function for this motion is:

`v=4t`

Now, we know that the velocity is defined by:

`v=(dx)/(dt)`

then the position can be obtained by:

`x=\int ^t_(t_0)vdt^(\prime)+x_0`

where t0 is some time and x0 is the position at that time; in this case we don't have an initial position for the particle then te position will be given by:

`\begin{gathered} x=\int ^t_(t_0)4t^(\prime)dt^(\prime)+x_0 \\ x=4((t^(\prime2))/(2))\vert^t_(t0)+x_0 \\ x=4((t^2)/(2)-(t^2_0)/(2))+x_0 \\ x=2t^2-2t^2_0+x_0 \end{gathered}`

Hence the position at any given time is given by:

`x=2t^2-2t^2_0+x_0`

Once we know the position function we can calculate the area undert the position time graph in the interval given:

`\begin{gathered} \int ^3_1(2t^2-2t^2_0+x_0)dt=((2)/(3)t^3+(x_0-2t^2_0)t)\vert^3_1 \\ =(2)/(3)(3^3-1^3)+(x_0-2t^2_0)(3^{}-1^{}) \\ =(2)/(3)(27-1)+(x_0-2t^2_0)(2) \\ =(2)/(3)(26)+(x_0-2t^2_0)(2) \\ =(52)/(3)+2(x_0-2t^2_0) \end{gathered}`

Therefore, in general, the area under the curve in the given interval is:

`(52)/(3)+2(x_0-2t^2_0)`

If we assume that the particle was at the origin at time t0=0 (this meas x0=0 as well), then the area under the position-time graph will be:

`(52)/(3)+2(0-2(0)^2)=(52)/(3)`

Note: No matter what the intial time and position is the expression we found will give us the correct answer.