Question

Pls help me i have to submit this due tomorrow :'D

Answer 1

`\textsf {a) As there is no change in the initial and final positions, the displacement is}\\\textsf {equal to 0.}`

b) Finding total distance :

Distance travelled from 0 s to 5 s :

• 1/2 x 5 x 30 = 75 m

Distance travelled from 5 s to 10 s :

• 5 x 30 = 150 m

Distance travelled from 10 s to 15 s :

• 150 + 1/2 x 5 x 5
• 150 + 12.5 = 162.5 m

Distance travelled from 15 s to 20 s :

• 36 x 5 + 1/2 x 5 x 4
• 180 + 10 = 190 m

Distance travelled from 20 s to 25 s :

• 45 x 5 + 1/2 x 5 x 5
• 225 + 12.5 = 237.5 m

Distance travelled from 25 s to 30 s :

• 40 x 5 + 1/2 x 10 x 5
• 200 + 25 = 225 m

Distance travelled from 30 s to 35 s :

• 20 x 5 + 1/2 x 20 x 5
• 100 + 50 = 150 m

Distance travelled from 35 s to 40 s :

• 1/2 x 20 x 5
• 50 m

Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50

Total = 1240 m

c) velocity at t = 15 s

• 36/15
• 12/5
• 2.4 m/s

d) average velocity

• 0 m/s (as displacement is equal to 0)

e) average speed

• 1240/40
• 31 m/s

f) Part d uses displacement whereas part e uses distance