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2 answers:
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Answer:
a) 27 m/s
b) 30 m/s
c) i) 3
ii) Deceleration
Step-by-step explanation:
The question is not complete, the correct question is given as:
The graph shows information about the speed of a vehicle during the final 50 seconds of a journey. At the start of the 50 seconds the speed is k metres per second. The distance travelled during the 50 seconds is 1.35 kilometres.
(a) Work out the average speed of the vehicle during the 50 seconds
(b) Work out the value of k.
(c) (i) Calculate the gradient of the graph in the final 10 seconds of the journey
(ii) Describe what this gradient represents
Answer:
The graph is attached. The total time = 50 seconds, total distance = 1.35 km = 1350 m
a) The average speed is the ratio of the total distance traveled to the total time taken to cover this distance. The average speed is given by the formula:
b) From the graph, the total distance covered is the area of the graph. The graph is made up of a rectangle and triangle, the area of the graph is equal to the sum of area of rectangle and area of triangle.
c) i) The gradient in the last 10 seconds is the ratio of change in speed to change in time
ii) Since the gradient is negative it means it is deceleration. That is in the in the last 10 seconds the vehicle decelerates at a rate of 3 m/s²
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Answer:
see below
Step-by-step explanation:
Setting x=0, you can find the y-intercept:
y = 4
Setting y = 0, you can find the x-intercept:
2x = 4
x = 2
The boundary line can be drawn through these intercept points, (0,4) and (2,0). Because the "equal to" case is not included, the boundary line will be dashed.
The y variable is on the open side of the comparison symbol (y>), so the shading is above the boundary line. That is where y-values are greater than those on the line.
