Question

The function \(s(t)\) describes the motion of a particle moving along a line. For each function, (a) find the velocity function of the particle at any time \(t \geq 0\), (b) identify the time interval(s) when the particle is moving in a positive direction, (c) identify the time interval(s) when the particle is moving in a negative direction, and (d) identify the time(s) when the particle changes its direction. $$ s(t)=t^{2}-7 t+10 $$

Short answer

The velocity function is \(v(t) = 2t - 7\). The particle is moving in a negative direction when \(t < 7/2\) and in a positive direction when \(t > 7/2\). The particle changes direction at \(t = 7/2\).

Step-by-step solution

Finding the Velocity Function

By calculating the derivative of the given motion function \(s(t)=t^{2}-7 t+10\), the velocity function can be obtained. The derivative is \( v(t) = 2t -7\), which describes the velocity of the particle at any time \(t\).

Determine Positive and Negative Direction

The particle moves in a positive direction when \(v(t)>0\) and in a negative direction when \(v(t)<0\). By setting \(v(t) = 0\), we find the critical value at \(t = 7/2\). Therefore, for \(t<7/2\), \(v(t)<0\), meaning the particle moves in a negative direction, and for \(t>7/2\), \(v(t)>0\), meaning the particle moves in a positive direction.

Identify Change of Direction

The particle changes direction at the points where \(v(t) = 0\), or when it does not exist. Since our \(v(t)\) exists for all \(t\), we only need to consider when \(v(t) = 0\). As calculated already, \(t = 7/2\) is the only time when the particle changes its direction.