Question

Consider a particle moving along a straight line, whose position as a function of time is given by \[ s(t)=\alpha t^2-\beta t+\gamma \] where \(\alpha=1\,\text{m s}^{-2}\), \(\beta=6\,\text{m s}^{-1}\) and \(\gamma=5\,\text{m}\). The average speed of the particle, in \(\text{m s}^{-1}\), from \(t=0\) to \(t=6\,\text{s}\) is:

• A. \(0\)
• B. \(12\)
• C. \(6\)
• D. \(3\)

Hint:

Average speed is based on total distance travelled, not displacement. Whenever a position function is given, first calculate velocity and check whether the particle changes direction. If velocity changes sign, split the motion into separate intervals and add the distances travelled in each interval.

Answer 1

The Correct Option is D

Concept: Average speed is defined as \[ \text{Average Speed} = \frac{\text{Total Distance Travelled}} {\text{Total Time Taken}} \] For motion along a straight line, distance travelled and displacement are generally different quantities. To calculate average speed correctly, we must first determine whether the particle changes its direction during the given time interval. The direction of motion is determined by the sign of velocity. Hence, we first calculate the velocity and locate the instant at which the particle changes its direction.

Step 1: Write the position function
Given, \[ s(t)=t^2-6t+5 \] The velocity is obtained by differentiating position with respect to time. \[ v=\frac{ds}{dt} \] \[ v=2t-6 \]

Step 2: Find the instant when the particle changes direction
A particle changes direction when its velocity becomes zero. Therefore, \[ 2t-6=0 \] \[ t=3\,\text{s} \] Thus the particle changes its direction at \[ \boxed{t=3\,\text{s}} \]

Step 3: Calculate the position at important instants
At \(t=0\), \[ s(0)=5 \] At \(t=3\), \[ s(3)=3^2-6(3)+5 \] \[ s(3)=9-18+5 \] \[ s(3)=-4 \] At \(t=6\), \[ s(6)=6^2-6(6)+5 \] \[ s(6)=36-36+5 \] \[ s(6)=5 \] Hence, \[ s(0)=5,\qquad s(3)=-4,\qquad s(6)=5 \]

Step 4: Determine the total distance travelled
Distance travelled from \(t=0\) to \(t=3\): \[ |5-(-4)| \] \[ =9\,\text{m} \] Distance travelled from \(t=3\) to \(t=6\): \[ |5-(-4)| \] \[ =9\,\text{m} \] Therefore, \[ \text{Total Distance} = 9+9 \] \[ =18\,\text{m} \]

Step 5: Calculate the average speed
Average speed is \[ \frac{\text{Total Distance}} {\text{Total Time}} \] \[ =\frac{18}{6} \] \[ =3\,\text{m s}^{-1} \]

Step 6: Write the final answer
Therefore, \[ \boxed{\text{Average Speed}=3\,\text{m s}^{-1}} \] Hence the correct option is \[ \boxed{\text{(D) }3} \]