Question:medium
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:
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:
Show 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.
Updated On: Jul 11, 2026
- \(0\)
- \(12\)
- \(6\)
- \(3\)
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Write the motion.
Position is $s(t) = \alpha t^2 - \beta t + \gamma$ with $\alpha = 1$, $\beta = 6$, $\gamma = 5$, so $s(t) = t^2 - 6t + 5$. We need the average speed from $t = 0$ to $t = 6\,\text{s}$.
Step 2: Find the velocity.
Differentiate: $v = \dfrac{ds}{dt} = 2t - 6$.
Step 3: Locate any turning point.
Velocity is zero when $2t - 6 = 0$, i.e. $t = 3\,\text{s}$. The particle reverses direction here, so distance and displacement differ.
Step 4: Positions at key times.
\[ s(0) = 5, \quad s(3) = 9 - 18 + 5 = -4, \quad s(6) = 36 - 36 + 5 = 5 \]
Step 5: Total path length.
From $0$ to $3$: $|{-4} - 5| = 9\,\text{m}$. From $3$ to $6$: $|5 - (-4)| = 9\,\text{m}$. Total distance $= 18\,\text{m}$.
Step 6: Average speed.
Divide distance by time:
\[ \text{Average speed} = \frac{18}{6} = 3\,\text{m s}^{-1} \]
\[ \boxed{3\,\text{m s}^{-1}} \]
Position is $s(t) = \alpha t^2 - \beta t + \gamma$ with $\alpha = 1$, $\beta = 6$, $\gamma = 5$, so $s(t) = t^2 - 6t + 5$. We need the average speed from $t = 0$ to $t = 6\,\text{s}$.
Step 2: Find the velocity.
Differentiate: $v = \dfrac{ds}{dt} = 2t - 6$.
Step 3: Locate any turning point.
Velocity is zero when $2t - 6 = 0$, i.e. $t = 3\,\text{s}$. The particle reverses direction here, so distance and displacement differ.
Step 4: Positions at key times.
\[ s(0) = 5, \quad s(3) = 9 - 18 + 5 = -4, \quad s(6) = 36 - 36 + 5 = 5 \]
Step 5: Total path length.
From $0$ to $3$: $|{-4} - 5| = 9\,\text{m}$. From $3$ to $6$: $|5 - (-4)| = 9\,\text{m}$. Total distance $= 18\,\text{m}$.
Step 6: Average speed.
Divide distance by time:
\[ \text{Average speed} = \frac{18}{6} = 3\,\text{m s}^{-1} \]
\[ \boxed{3\,\text{m s}^{-1}} \]
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