Question

Umm just want u all to check my work if it's right or not

Answer 1

Part A and part B are correct

Answer 2

for part C, we're looking at the deceleration stage

\(\Delta\)v = Vf - Vi

you got that backwards

The final speed is 0
The initial speed in the deceleration stage is 12 m/s

so Vf = 0
Vi = 12

so recalculate \(\Delta v\)

and the final units you should get are
m/s / s = m/s * 1/s = m/s^2

Answer 3

Part D is also wrong

We need to calculate the distance travelled in each stage

Or
since we have a graph of speed vs time
since distance = speed * time

the area under the graph is the total distance travelled

Answer 4

So we know x = 20


We have a triangle with a base of 20 and height of 12

We have a rectangle with a length of 5 (because 25-20 = 5)
and a width of 12

We have a triangle with a base of 35 and height of 12

Can you find the area of each shape and add them all up together? That is going to be the total distance travelled.

Answer 5

@villy

Answer 6

@AZ can u can u check the next ones

Answer 7

part A is wrong

they travelled 4 meters and then they travelled 4 meters back

so the total distance is 4 + 4

Answer 8

the second part is correct :)

Answer 9

@AZ

Answer 10

on the last part

d = 60 m not 60 m/s

Answer 11

\[d = (6 m/s) * (10 s) = 60 (ms/s) = 60 (m)\]

Answer 12

Part b) is wrong

6 m/s was her speed at the end but it was constantly changing so you can't use 6 m/s

you have to find the average velocity first

average velocity = (1+5)/2

and then use that
distance travelled = velocity * time

Answer 13

alternatively, if you remember the distance travelled by the cyclist is the area under the graph

you can break the area underneath into two shapes

a rectangle with a width of 1 and length of 10

and a triangle with a base of 10 and a height of 5

find the area of both shapes and add them

Answer 14

alternatively, if you remember the distance travelled by the cyclist is the area under the graph

@AZ i think this can be so confusable to @villy bc. the distance traveled by a cyclist not can be an area like you expressed it above ?
@Shadow opinion in this way pls ? ty

Answer 15

I think you guys got this. AZ seems on point.

Answer 16

It's a graph of speed by time

and distance = speed * time

so the area under the graph gives us the distance

`The area under the line on a velocity-time graph is equal to the displacement of the object. If the shape of the graph can be broken into simple geometric shapes, the total area under the line can be calculated by adding the areas of those shapes. The area under a speed-time graph is the distance.`

https://isaacphysics.org/pages/gcse_ch2_12_text#:~:text=The%20area%20under%20the%20line,the%20displacement%20of%20the%20object.&text=If%20the%20shape%20of%20the,time%20graph%20is%20the%20distance.

Answer 17

ok i accept this all but how you explain to a student that a cyclist when go on a distance of 300 m in a line (way) that the distance graphed not is this line - really is the area - this is so absurd