# Intuition of Distance covered when accelerating [duplicate]

When you're moving at $$5$$ m/s for $$1$$ second, you have traveled $$5$$ m.

When you're moving at $$5$$ m/s (initial velocity) and you accelerate $$2$$ m/s for $$1$$ second, you have traveled $$5$$ m + $$1$$ m (distance traveled because of acceleration).

But does this hold true in real life? If I was to test this out, would the distance traveled because of acceleration equal $$1$$ m?

• I'm not sure what you're asking or of the motivation. Are you asking if kinematics holds true? Commented Jun 6, 2019 at 22:20
• Kind of. I learned that the distance traveled when accelerating is 1/2(b*h), but I wonder if that holds true in real life too. Commented Jun 6, 2019 at 22:29
• The short answer is "yes". To understand this better, you should make "in real life" more concrete: design an experiment to compare the baseline and the variant cases. At that point the "in real life" versus theory crystallises into "what factors do I need to take into account in the experiment design to ensure that we really satisfy the premises of the model". Commented Jun 7, 2019 at 5:48
• Possible duplicates: physics.stackexchange.com/q/89590/2451 and links therein. Commented Jun 7, 2019 at 10:58
• This Q has a distinguishing feature which is that it needs to be addressed with not only reference to a comprehension of measurement procedure and curve fitting, and also a NB that real life motors and such are not going to have the same torque at different speeds so it's hard to know exactly how long it really goes at which speed. Commented Jun 21, 2019 at 0:26

Let $$y$$-axis shows speed, and $$x$$-axis represents time:
The $$\color{red}{\mathrm {red\ line}}$$ is the graph for the first moving with $$\color{red}{constant}$$ speed, the $$\color{blue}{\mathrm {blue\ one}}$$ for your second (accelerated) moving with $$\color{blue}{increasing}$$ speed.
The area under them is the traveled distance, so you may see, that for the accelerated moving it's the sum of traveled distance for moving with constant speed plus area of the triangle over the rectangle, which is always $$at/2$$, giving in your case $$2\cdot 1 / 2 = 2\cdot t / 2 = t=1$$.
So in the more general case, when time may be other than $$1$$ s, but the acceleration is still $$2\,\mathrm{ms^{-2}}$$, it is true, but only numerically, that $$s_2=s_1+t$$.