Introductory physics: first second of free fall While doing my physics homework I encountered an acceleration problem that I've been unable to solve.
Here, is the data table of the objects motion. At 1 second, why does the "Dist-meters" slot of the table have a value of 4.9m instead of 9.8m? Shouldn't the distance be rather 9.8m since the velocity at 1 second is 9.8m/s (meaning at 1 second, the object traveled 9.8 meters).
What am I doing wrong? I encountered this problem because I tried to understand the concept conceptually, before memorizing the formula. I also realized this with another free fall problem .I made a data table of the car during the 4 second interval; after searching on the web for an hour, I finally arrived at the correct answer when I changed the distance at 1 sec to 5m (half the velocity at that second) from 10m.
What is this??  
 A: You don't need calculus to understand this and I think you are right to be trying to gain a deeper understanding than just memorizing some formulas.
During that first second the body accelerates - it starts with 0 velocity and gains linearly giving 9.8 at the end of the first second, so at that point, it hasn't been moving at 9.8m/s for a second, it has been moving with a changing velocity. The most intuitive way to measure the distance traveled is to just take the average velocity for that first second and multiply it by that interval - $(v-v_0)t/2$, which in your case is $(9.8-0)*1/2)$.
Calculus would be useful for an acceleration that changes with time or a more rigorous derivation.
A: Intuitively, the body spends some time at every velocity between 0 and 9.8 m/s during the first second. From the formula distance = speed * time, if we call that time interval dt and add up all the contributions (using integral calculus) the answer is 4.9 m.
A: Start with acceleration, which we assume to be a constant $g$. Also assuming 'up' is the positive spacial direction i.e. $g$ is $(-)ve$:
   $$a(t) = -g = -9.81 \,ms^{-2}$$
Integrate once to get velocity:
   $$\dot{a} =v(t) = \int_0^t -g dt = -gt +v_0$$
Integrate again to get the distance:
   $$\ddot a = x(t) = \int_0^t(-gt + v_0)\, dt = -\frac{1}{2}gt^2 +v_0t +x_0$$
Now assuming that the reference $x_0 = 0$, i.e. where the object is at rest and you don't give it any initial velocity $v_0 = 0$, after time $t=1\, s$ has passed the position of the object would be at: 
$$-\frac{g}{2} = -4.9 \,m$$
i.e. $4.9\,m$ below where it started.
Calculus is your friend :)
A: Draw a graph with time along the horizontal and velocity up the vertical. 
Let's start with an object in motion at constant velocity. Its motion on the graph will be represented by a horizontal line at some distance from the y=0 axis. After some period of time, it will have covered a distance equal to velocity x time. That distance will be represented on the graph as the area under the line. Consider how increasing or decreasing the speed or time would affect distance traveled, and how it would affect the area under the line - thus we can make the intuitive connection between distance traveled and area on our speed-time graph.
Now consider an object released into free fall. It will have velocity zero at time zero. Since we are talking about Earth surface gravity, we've established that at time 1 second, velocity will be 9.8m/s, and since acceleration is constant, velocity will increase uniformly with time. On the graph, that will be represented by a straight (sloped) line intersecting the points (time=0,velocity=0) and (time=1sec,velocity=9.8m/s). This will appear on the graph as a triangle. As with the previous example, distance traveled will be represented by the area under the line, which we calculate as height x width / 2 (area of a right triangle), thus we get 4.9m. 
