Help question 6

Distance = Speed × Time

Distance = (20 m/s) × (10 s)

Distance = 200 m

The distance travelled during the first 10 seconds is 200 m.

Acceleration (a) = (Final Velocity (v) - Initial Velocity (u)) / Time (t)

Initial Velocity (u) = 20 m/s
Final Velocity (v) = 30 m/s
Time (t) = 5 s

a = (30 m/s - 20 m/s) / (5 s)
a = (10 m/s) / (5 s)
a = 2 m/s²

The acceleration during the period of uniform acceleration is 2 m/s².

Distance (s) = ut + (1/2)at²
Where u = initial velocity, t = time, a = acceleration.

u = 20 m/s
t = 5 s
a = 2 m/s²

s = (20 m/s) × (5 s) + (1/2) × (2 m/s²) × (5 s)²
s = 100 m + (1 m/s²) × (25 s²)
s = 100 m + 25 m
s = 125 m

The distance travelled during the period of uniform acceleration is 125 m.

Total Distance = Distance (Part 1) + Distance (Part 2) + Distance (Part 3)

Distance (Part 1) = 200 m (from part a)
Distance (Part 2) = 125 m (from part c)
Distance (Part 3) = Speed × Time = (30 m/s) × (15 s) = 450 m

Total Distance = 200 m + 125 m + 450 m
Total Distance = 775 m

The total distance travelled is 775 m.