Answer:
a) 4.50m/s²
b) v= 21
c) 11.25m/s
Step-by-step explanation:
a) rate of change of speed in first 2 seconds
= gradient of graph from t=0s to t=2s
The two points are (0,0) and (2,9).
[tex]gradient = \frac{y1 - y2}{x1 - x2} [/tex]
Rate of change of speed
[tex] = \frac{9 - 0}{2 - 0} \\ = \frac{9}{2} \\ = 4.50m/ {s}^{2} [/tex]
Total distance= area under graph
Distance travelled for first 6s
= area of traingle +area of rectangle
= ½(2)(9) +(6-2)(9)
= 9 + 4(9)
= 9 +36
= 45m
Distance travelled from t=6s to t=12s
= 135 -45
= 90m
Area under graph from t=6s to t=12s =90
Area of trapezium= 90
½(A +B)(h)= 90
½(9+ v)(6)= 90
3(9 +v)= 90 (simplify)
9 +v= 30 (÷3 throughout)
v= 30 -9
v= 21
c) Average speed
= total distance ÷total time
= 135 ÷12
= 11.25 m/s
