HELP IN MATHS

[Lana19984]

http://www.thamerinternationalschool.com/uploads/files/0580_w09_qp_21.pdf
Would anyone please help me with question 21 part b ? i dont understand how the answer is 960 m
PLEASE HELP

 

[notnek01]

http://www.thamerinternationalschool.com/uploads/files/0580_w09_qp_21.pdf
Would anyone please help me with question 21 part b ? i dont understand how the answer is 960 m
PLEASE HELP

Imagine the truck line isn't there - it's probably confusing you.

The distance travelled by the car is the area under the line that goes to the x-axis (it doesn't stop at the truck line, so ignore that line).

The car is travelling faster than the truck between 15 and 55 seconds so you need to find the area under the line between those two times.

 

[Lana19984]

Imagine the truck line isn't there - it's probably confusing you.

The distance travelled by the car is the area under the line that goes to the x-axis (it doesn't stop at the truck line, so ignore that line).

The car is travelling faster than the truck between 15 and 55 seconds so you need to find the area under the line between those two times.

but why do i ignore it ?

 

[notnek01]

but why do i ignore it ?

It's hard to know your understanding.

Can you try to explain to me what you think the method should be? And then I will correct you.

Maybe post your working?

 

[Lana19984]

It's hard to know your understanding.

Can you try to explain to me what you think the method should be? And then I will correct you.

Maybe post your working?

i though basically its just the triangle above the line

 

[masterex567]

http://www.thamerinternationalschool.com/uploads/files/0580_w09_qp_21.pdf
Would anyone please help me with question 21 part b ? i dont understand how the answer is 960 m
PLEASE HELP

the truck is moving at a constant speed of 12 m/s.
The only time that the car travels faster is between 15 to 55 seconds, where it's speed exceeds the one of the truck as shown by graph.

So calculate the distance moved by car between 15 to 55 seconds, and increased from a speed of 12 to 36 between those seconds.
to do this we need to find area under the graph. this area forms two triangles: one from 15 to 45 seconds. another from 45 to 55 seconds. so calculate the area of these triangles using the change in time as base, and change in speed as height.

=1/2 * (45-15) * (36 - 12) + 1/2 * (55-45) * (36 - 12) which is 480.
NOW, remember we have to find distance UNDER the graph. which includes the small rectangle that is below the truck line to the graph axis.
this is (55-15) * 12 = 480

Hence add the two, 480 + 480 which gives 960m

 

[S.sarraf]

Heyy does anyone have maths paper 4 2015 of March series ? My exam is tmmr and I need it badly . Thank you in advance . You. Can mail me on [email protected] . May god bless you .