plz plz plz ans this ques
Attachments:
Answers
Answered by
1
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
4ans is -8.5m/s
5ans is 21 m
can u do it plsssssss
Ok! Plz wait
Oh I'm Sry! Yeah, You have to take the y coordinate corresponding to point C as 17. Then for 4) Average velocity = (Final velocity + Initial velocity) / 2 = (-17 +0) / 2 = -8.5 ms^-1 And for 5) (17+25)/2 = 21m
thankuuu so much
Anytime!
Similar questions