The correct answers for the given problem are: (a) The distance travelled on the cycle in the first 14 seconds is 96 m and (b)The acceleration in the first 4 seconds is 2 m/s².
The area under a graph can be found by dividing the graph into equal intervals of very small width and then find the area of the rectangle for each intervals. After summing all the areas of rectangles the area under the graph can be approximated.
The velocity time graph given in the problem can be used to find the distance travelled.
The distance travelled in first 14 seconds is equal to the area of the graph from t = 0 to t = 14.
This area consists of a triangle of height h = 8 and base b = 4 and a rectangle of length l = 8 and width b = (14 - 4) = 10.
Now, the total area is given as,
1/2 × b × h + l × b
= 1/2 × 4 × 8 + 8 × 10
= 16 + 80
= 96
(b) The acceleration is given as the rate change of velocity with respect to time.
The acceleration in first 4 seconds is given as,
a = (8 - 0)/4
= 2
Hence, the distance travelled in first 14 seconds and acceleration in first 4 seconds is given as 96 m and 2 m/s² respectively.
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