Determinatio/Goal setting
Answer:
Explanation:
s = 50 t^2
Differentiate both sides with respect to t
ds/dt = 50 x 2 t
ds/dt = 100 t
Substitute t = 10 second
ds/dt = 100 x 10
ds/dt = 1000 m/s
thus, the rate of change of elevation is 1000 m/s.
First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s
Answer:
Distance is 500 m, displacement is 0
Explanation:
Distance and displacement are defined in two different ways:
- Distance is the total length of the path covered by an object in motion - so it depends on the path taken. In this problem, the distance travelled by the car corresponds to the length of one lap, which is the length of the track, so 500 m
- Displacement is the distance in a straight line between the final point and the initial point of the motion. This means that displacement does not depend on the path taken, but only on the starting and ending point of the motion. In this problem, the car completes one lap, so the final position of the car is equal to its starting position - therefore the displacement is zero, since the distance between these two points is zero.
The stomach is above the waist, below the waist is your, yunno. the stomach and bladder sit right on top of the waist, hope this helps, have an amazing day:)
