# An object travels North at 7 m/s for 5 s and then travels South at 9 m/s for  2 s. What are the object's average speed and velocity?

May 13, 2016

${\vec{V}}_{\text{net"=2.43m/s->"towards north}}$

${V}_{\text{avg}} \approx 7.6 \frac{m}{s}$

#### Explanation:

For 1st 5s the velocity of the object towards North be
${\vec{V}}_{\text{north}} = 7 \hat{j} \frac{m}{s}$

For 1st 5s the displacement of the object towards North be
${\vec{d}}_{\text{north}} = 7 \hat{j} \frac{m}{s} \times 5 s = 35 \hat{j} m$

For 2nd 2s the velocity of the object towards Southbe ${\vec{V}}_{\text{south}} = - 9 \hat{j} \frac{m}{s}$

For 2nd 2s the displacement of the object towards Southbe ${\vec{d}}_{\text{south}} = - 9 \hat{j} \frac{m}{s} \times 2 s = - 18 \hat{j} m$

Net displacement during t=(5+2)=7s is
${\vec{d}}_{\text{net"=35hatjm-18hatjm=17hatjm->"towards north}}$

Hence the Net Velocity
${\vec{V}}_{\text{net"=vecd_"net"/t=17/7hatjm=2.43m/s->"towards north}}$

Total distance covered during 7 s
${d}_{\text{total"=|vecd_"north"|+|vecd_"south}} | = | 35 \hat{j} m | + | - 18 \hat{j} m | = 53 m$

Hence Average speed
${V}_{\text{avg" = d_"total}} / t = \frac{53}{7} \frac{m}{s} \approx 7.6 \frac{m}{s}$