plz plz plz ans this ques
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1) In a v-t graph the gradient shows the acceleration. (Used two coordinates)
∴ a from A-B = (25 - 5)/(3-1) = 10 ms^-2 [gradient = y axis/x axis]
2) Similarly,
a from B-C = (10-25)/(5-3) = - 7.5 ms^-2 (Here a negative gradient means deceleration)
3) In a v-t graph the area shows the displacement,
∴ s = 1/2 (3x25) [Area of the triangle ABE]
= 37.5 m
4) Average velocity = Total displacement / Total time
= {1/2(2*15.5)} / 2
= 15.5 / 2
= 7.75 ms^-1
5) Area of the trapezium = 1/2(15.5 + 25) * 1
= 20.25 m
∴ a from A-B = (25 - 5)/(3-1) = 10 ms^-2 [gradient = y axis/x axis]
2) Similarly,
a from B-C = (10-25)/(5-3) = - 7.5 ms^-2 (Here a negative gradient means deceleration)
3) In a v-t graph the area shows the displacement,
∴ s = 1/2 (3x25) [Area of the triangle ABE]
= 37.5 m
4) Average velocity = Total displacement / Total time
= {1/2(2*15.5)} / 2
= 15.5 / 2
= 7.75 ms^-1
5) Area of the trapezium = 1/2(15.5 + 25) * 1
= 20.25 m
Veer1111:
can u plz do 4,5 part again
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