In a distance time graph, when the x and y values are positive (in first quadrant), the runner is moving forward.
While, if the distance value ( in the y axis) is negative, the runner is moving backwards ( towards the start) .
Hope it helps :)
y = 75.9 m
Explanation:
y = -(1/2)gt^2 + v0yt + y0
If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.
y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)
= -44.1 m + 120 m
= 75.9
B. velocity at position x, velocity at position x=0, position x, and the original position
In the equation
=
+2 a x (x - x₀)
= velocity at position "x"
= velocity at position "x = 0 "
x = final position
= initial position of the object at the start of the motion
Answer:
Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.
Explanation:
Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.
Answer:
F = 3.20 N
Explanation:
Given:
Work done by child = 80.2 j
Distance that the car moves = 25.0 m
We need to find the force acting on the car.
Solution:
Using work done formula as.

Where:
W = Work done by any object.
F = Force (push or pull)
d = distance that the object moves.
Substitute
in work done formula.


F = 3.20 N
Therefore, force acting on the car F = 3.20 N