# 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??

• 9.8 m/s^2 is not a velocity. Jan 31, 2016 at 1:47
• it should be 9.8m/s. my mistake. Jan 31, 2016 at 1:50
• @ACuriousMind now that I've corrected the mistake, do you know the answer? Jan 31, 2016 at 1:54
• Think about the fact that the velocity is not constant during that one second. Jan 31, 2016 at 1:56
• I didn't know the velocity was not constant, and that does not explain why the distance =*half* of the velocity instead of some other constant. This is also my first physics class. Jan 31, 2016 at 2:00

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.

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.

• thank you for the explanation. I've been extremely perturbed about this since yesterday. I don't fully understand the answer because I've only studied calc up to derivatives with little integration. Do you know why this was not explained in my physics book (it's algebra based)? Henceforth, should I memorize the formulas instead of trying to understand the concepts? Jan 31, 2016 at 2:35

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.