Answer:
a) The acceleration in the first 20 seconds is 0.35 m/s²
b) The distance between the two stops is 945 m
c) The average velocity of the bus during the journey is approximately 5.9 m/s
Step-by-step explanation:
a) From the graph, we have;
The velocity of the bus at time t₁ = 0 s, v₀ = 0 m/s
The velocity of the bus at time t₂ = 20 s, v₂₀ = 7 m/s
The uniform acceleration, 'a', is given by the following equation;
a = dv/dt
Therefore, the acceleration in the first 20 seconds is given as follows;
a = (v₂₀ - v₀)/(t₂ - t₁) = (7 m/s - 0 m/s)/(20 s - 0 s) = 0.35 m/s²
The acceleration in the first 20 seconds, a = 0.35 m/s²
b) The two stops are at the start of the graph (0,0) and the point (160, 0), which is the distance traveled during the journey
Therefore, a way to find the distance between the two stops is by finding the area under the velocity time graph
Here the area consists of 2 triangles and one rectangle
The area of the triangle on the left, A₁ = 1/2 × 20 s × 7 m/s = 70 m
The area of the middle rectangle, A₂ =(130 s - 20 s) × 7 m/s = 770 m
The area of the triangle on the right, A₃ = 1/2 × (160 s - 130 s) × 7 m/s = 105 m
The area under the velocity time graph, A = A₁ + A₂ + A₃ = 70 m + 770 m + 105 m = 945 m
The distance between the two stops = The area under the velocity time graph = 945 m
c) The average velocity of the bus during the journey, [tex]v_{avg}[/tex] = (The distance between the two stops)/(The time taken to cover the distance)
∴ [tex]v_{avg}[/tex] = (945 m)/(160 s) = 5.90625 m/s
The average velocity, [tex]v_{avg}[/tex] ≈ 5.9 m/s (rounded to 1 decimal place (d. p.)
