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 $\alpha=1\ ext{ms}^{-2}, \beta=6\ ext{ms}^{-1}$ and $\gamma=5\ ext{m}$. The average speed of the particle, in $ ext{ms}^{-1}$ from $t=0$ to $t=6\ ext{s}$ is:

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 $\alpha=1\ 	ext{ms}^{-2}, \beta=6\ 	ext{ms}^{-1}$ and $\gamma=5\ 	ext{m}$. The average speed of the particle, in $	ext{ms}^{-1}$ from $t=0$ to $t=6\ 	ext{s}$ is:

Accepted Answer

D. 1. **Set up the position equation** by substituting the given values into $s(t)=\alpha t^2-\beta t+\gamma$.

Since $\alpha=1\ 	ext{ms}^{-2}$, $\beta=6\ 	ext{ms}^{-1}$, $\gamma=5\ 	ext{m}$, we get

$$s(t)=t^2-6t+5$$

2. **Find the velocity** by differentiating position with respect to time, since velocity is the rate of change of position.

$$v(t)=\frac{ds}{dt}=2t-6$$

3. **Find when the velocity becomes zero**, because that is when the particle reverses direction, which matters for calculating total distance (average speed depends on distance, not displacement).

Setting $v=0$:

$$2t-6=0 \Rightarrow t=3\ 	ext{s}$$

4. **Determine the direction of motion in each interval.** For $0<t<3$, velocity is negative (particle moves backward); for $3<t<6$, velocity is positive (particle moves forward).

5. **Calculate the distance covered in each interval using the area under the $v$-$t$ graph** (a triangle in each interval).

For $t=0$ to $t=3$:

$$d_1=\frac{1}{2}	imes 3	imes 6=9\ 	ext{m}$$

For $t=3$ to $t=6$:

$$d_2=\frac{1}{2}	imes 3	imes 6=9\ 	ext{m}$$

6. **Add both distances to get the total distance**, since average speed uses total distance, not net displacement.

$$d=d_1+d_2=9+9=18\ 	ext{m}$$

7. **Apply the average speed formula**: average speed equals total distance divided by total time.

$$v_{avg}=\frac{d}{\Delta t}=\frac{18}{6}=3\ 	ext{ms}^{-1}$$

Hence, the answer is **D. 3**.