In the physics world, velocity (v), position (x), acceleration (a) and time (t) are the four key ingredients in solving equations of motion. You may get the acceleration, initial velocity (v0) and elapsed time of a particle and have to solve for the final velocity (vf). A variety of other permutations applicable to countless real-world scenarios are possible. These concepts appear in four essential equations:
(1. x=v_0t+frac{1}{2} at^22. v_f^2=v_0^2+2ax3. v_f=v_0+at4. x=frac{v_0+v_f}{2}t)
These equations are useful in calculating the speed (equivalent to velocity for present purposes) of a particle moving with constant acceleration at the moment it strikes an unyielding object, such as the ground or a solid wall. In other words, you can use them to calculate impact speed, or in terms of the above variables, vf.
Step 1: Assess Your Variables
If your problem involves an object falling from rest under the influence of gravity, then v0 = 0 and a = 9.8 m/s2 and you need only know the time t or the distance fallen x to proceed (see Step 2). If, on the other hand, you may get the value of the acceleration a for a car traveling horizontally over a given distance x or for a given time t, requiring you to solve an intermediate problem before determining vf (see Step 3).
Step 2: A Falling Object
If you know an object dropped from a rooftop has been falling for 3.7 seconds, how fast is it going?
From equation 3 above, you know that:
(v_f=0+(9.8)(3.7)=36.26text{ m/s})
If you are not given the time but know that the object has fallen 80 meters (about 260 feet, or 25 stories), you would use equation 2 instead:
(v_f^2=0+2(9.8)(80)=1568v_f=sqrt{1568}=39.6text{ m/s})
You’re done!
Step 3: A Speeding Car
Say you know that a car that started from a standstill has been accelerating at 5.0 m/s for 400 meters (about a quarter of a mile) before driving through a large piece of paper set up for a celebratory display. From equation 1 above:
(400=0+frac{1}{2}(5)t^2=2.5t^2160=t^2t=12.65text{ seconds})
From here, you can use equation 3 to find vf:
(v_f=0+(5)(12.65)=63.25text{ m/s})
Tip
Always use an equation first for which there is only one unknown, which is not necessarily one that contains the variable of ultimate interest.
