The velocity function is $v (t) = - t^{2} + 4 t - 3$ $v(t)=-t^2+4t-3$ for a particle moving along a line. Find the displacement of the particle during the time interval [0,5]?

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Answer 1 · 2015-07-26T08:38:45

The problem is illustrated below.

Here, the velocity of the particle is expressed as a function of time as,

$v (t) = - t^{2} + 4 t - 3$ $v(t) = - t^2 + 4t - 3$

If $r (t)$ $r(t)$ is the displacement function, it is given as,

$r (t) = \int_{t_{0}}^{t} v (t) \cdot d t$ $r(t) = int_(t""_0)^t v(t)*dt$

According to the conditions of the problem, $t_{0} = 0$ $t""_0 = 0$ and $t = 5$ $t = 5$ .

Thus, the expression becomes,

$r (t) = \int_{0}^{5} (- t^{2} + 4 t - 3) \cdot d t$ $r(t) = int_0^5 (-t^2 + 4t - 3)*dt$

$\Rightarrow r (t) = (- \frac{t^{3}}{3} + 2 t^{2} - 3 t)$ $implies r(t) = ( -t^3/3 + 2t^2 -3t)$ under the limits $[0, 5]$ $[0,5]$

Thus, $r = - \frac{125}{3} + 50 - 15$ $r = -125/3 + 50 - 15$
The units need to be put.

The velocity function is v(t)=-t^2+4t-3v(t)=−t2+4t−3v(t)=-t^2+4t-3 for a particle moving along a line. Find the displacement of the particle during the time interval [0,5]?