A ball is thrown upwards and returns to the throwers hand.
Air resistance is ignored and the upwards direction is taken as positive.
Circle the correct values for the acceleration of the ball:
Going up : g / 0 / -g
At max height: g / 0 / -g
Going down: g / 0 / -g
2 model cars are at opposite ends of a 2000mm track.
When timing starts A is moving with a uniform velocity of 50mms-1 towards B.
B starts from rest when timing starts, then accelerates uniformly at 4mms-2 for 10s and then continues towards A with the uniform velocity it has reached.
Calculate the distance from the initial position of A the two cars meet.
Equations of Motion
The Attempt at a Solution
On the way up the ball is moving in the opposite direction to the force of gravity so g will be -ve as the ball is deccelerating
Then on the way down the opposite is true so g will be +ve.
However at the max height I know the acceleration cannot be 0 as the ball is still being pulled by the force of gravity however I can't decide whether it would be +ve or -ve g.
I think it must be -ve as at the max height it then immediately starts moving downwards so it must have just experienced an acceleration to make it move downwards?
Or should that all be reversed because of the upwards is positive clause?
I calculated the distance travelled by both A and B during the first 10 seconds and got:
A travels 500mm
B travels 200mm
Then there is 1300mm of track left.
A is travelling at 50mms-1 as it is travelling at constant uniform velocity.
I then calculated that B is travelling at 40mms-1 in the opposite direction.
Then I looked at it from the point of view of A.
So A is travelling effectively at 90mms-1 over 1300mm to meet with B.
This then allows me to calculate that that time taken to cover this distance would be 24.444... seconds.
Then I add on the 10 seconds I used earlier.
And use s = 0.5 * (u + v) * t with the values for A to find a distance of 1222.222... mm away from A's starting point?