This is for Question 3 part 2. My approach was to find the gradient as they both shared a common point X which is the speed. For the Gradient of the line from 62-48 Seconds I got the coordinates (48, 12.6) and (62,X). Then for the Gradient of the line from 70-62 I got the coordinates (62,X) and (70,0). Then I put the gradient of the first line equal to Two times the gradient of the second line. Getting a value of a=-0.7. Can anyone explain why we cannot use this approach?
you should multiply the gradient of first line by 2 and equate it with the gradient of second line as the second line's gradient is two times the first lines's so if we multiply the gradient of first line by 2 then it will equal to the gradient of second line
why do we have to take deceleration as negative, is it not only the case when its freefall as when its going up a is negative and when the object is coming down a is positive?
We take acceleration as negative when body becomes slow down. In 3rd question there is a cofficient of friction between plane and particle which means plane is rough and particle is going to be slow down
