Question

Help with #2 on physics, the one on the bottom

Answer 1

Part (a)

For interval 0 to 5 s, the velocity of the particle increases as the time taken by particle increases.

For interval 5 s to 10 s, the velocity of particle remains constant.

For interval 10 s to 15 s, the velocity of the particle decreases.

For interval 15 s to 20 s, the velocity of particle remains constant.

For interval 20 s to 25 s, the of particle decreases with increase in time.

Part (b)

The distance travelled by the particle for first 15 seconds is equal to the area of the graph which is calculated as,

`\begin{gathered} d=(1)/(2)(30\text{ m/s)(5 s-0 s)+}(30\text{ m/s)(10 s-5 s)} \\ \text{+}(1)/(2)(30\text{ m/s-20 m/s)(15 s-10 s)}+(20\text{ m/s)}(15\text{ s-10 s)} \\ =75\text{ m+150 m+}25\text{ m+}100\text{ m} \\ =350\text{ m} \end{gathered}`

Therefore, the total distance covered by the particle for first 15 seconds is 350 m.

Part (c)

The distance covered by particle from 15 s to 25 s is calculated as.

`\begin{gathered} d=(20\text{ m/s)(20 s-15 s)+}(1)/(2)(20\text{ m/s)(25 s-20 s)} \\ =(20\text{ m/s)(5 s)+(10 m/s)(5 s)} \\ =100\text{ m+50 m} \\ =150\text{ m} \end{gathered}`

Therefore, the distance covered by particle in interval 15 s to 25 s is 150 m.