The graph below describes the journey of a bus between two stops. a) Work out the
acceleration, in m/s2,
in the first 20 s.
b) Work out the distance
between the two stops.
c) Work out the average
speed of the bus during
the journey.
Give
your answer to 1 d. p.​

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, = (The distance between the two stops)/(The time taken to cover the distance)

∴ = (945 m)/(160 s) = 5.90625 m/s

The average velocity, ≈ 5.9 m/s (rounded to 1 decimal place (d. p.)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x