An object travels North at 1 m/s for 8 s and then travels South at 7 m/s for  5 s. What are the object's average speed and velocity?

Jul 4, 2016

Average Speed = $3.31 m {s}^{-} 1$
Average Velocity = $- 2.08 m {s}^{-} 1$ North ($2.08 m {s}^{-} 1$ South)

Explanation:

Speed does not rely on the direction of movement so:
$\text{Average speed" = "Displacement (total distance travelled)"/"Time}$

Displacement ($s$) = Displacement 1 (${s}_{1}$) + Displacement 2 (${s}_{2}$)
$s = \text{speed} \times \Delta t$
${s}_{1} = 1 m {s}^{-} 1 \times 8 s = 8 m$
${s}_{2} = 7 m {s}^{-} 1 \times 5 s = 35 m$

${s}_{\text{total}} = 8 m + 35 m = 43 m$
Time (total) = $8 s + 5 s = 13 s$

$\text{Average Speed} = \frac{s}{t} = \frac{43 m}{13 s} = 3.31 m {s}^{-} 1$

Velocity is reliant on the direction of travel so:

$\text{Velocity ("v")"=("Final distance from start ("d_"total"")")/("Time ("t")}$

Here we use velocity ($v$) with a direction so that movement North is positive (and movement South is negative)
$\Delta d = v \times \Delta t$
${d}_{1} = 1 m {s}^{-} 1 \times 8 s = 8 m$
${d}_{2} = - 7 m {s}^{-} 1 \times 5 s = - 35 m$

${d}_{\text{total}} = {d}_{1} + {d}_{2} = 8 m + \left(- 35 m\right) = - 27 m$

v=(d_"total")/(t_total) = (-27m)/(13s)=-2.08ms^-1 " North"
You could also say the velocity is $2.08 m {s}^{-} 1 S o u t h$. Since both are equivalent, both are correct, as long as you give the correct direction to go with the answer.