# Thread: Determining the displacement of a particle from velocity function by integration

1. ## Determining the displacement of a particle from velocity function by integration

A particle moves along a straight line from a fixed point $A$.Its velocity,
$v m s^-1,t s$ after leaving point $A$ is given by $v=3 + 2t-t^2$

a)Find the total distance travelled by the particle between $t=2$ and $t=4$

i get,

total distance travelled by the particle between $t=2$and $t= 3$

$=\int_{2}^{3} v dt$

$=\int_{2}^{3} (3 + 2t - t^2) dt$

$=\left[3t+t^2-\frac{t^3}{3} \right]_{2}^{3}$

$=\left[3(3) + 3^2 - \frac{3^3}{3} \right] - \left[3(2) + 2^2 - \frac{2^3}{3}$

$=1\frac{2}{3} m$

total distance travelled by the particle between $t = 3$ and $t = 4$

$=\int_{3}^{4} v dt$

$=\int_{3}^{4} (3 + 2t - t^2) dt$

$=\left[3t + t^2 - \frac{t^3}{3} \right]_{3}^{4}$

$=\left[3(4) + 4^2 - \frac{4^3}{3} \right] - \left[3(3) + 3^2 - \frac{3^3}{3} \right]$

$=6\frac{2}{3} - 9$

$=-2\frac{1}{3}$

$=2\frac{1}{3} m$

therefore,total distance travelled by the particle between $t=2$ and $t = 4$

$=1\frac{2}{3} + 2\frac{1}{3}$

$= 4m$

is that correct?

b)determine the maximum displacement of the particle from point $A$ before it reverses the direction of motion..

how can i find?

2. Originally Posted by faraday
A particle moves along a straight line from a fixed point $A$.Its velocity,
$v m s^-1,t s$ after leaving point $A$ is given by $v=3 + 2t-t^2$

a)Find the total distance travelled by the particle between $t=2$ and $t=4$

i get,

total distance travelled by the particle between $t=2$and $t= 3$

....
is that correct?

b)determine the maximum displacement of the particle from point $A$ before it reverses the direction of motion..

how can i find?
1. Your result looks good.

2. You should mention why you split the integral at t = 3.

3. And if you have done that you know how to answer question b): Remember the particle is moving on a straight line, so what speed has the particle when it starts to return?

3. mean that $\frac{ds}{dt}=0$ ?