In the case of accelerated motion there are more formulas. In addition to velocity (v), time )t) and displacement (s), there is acceleration (a) and initial velocity(u):
$$\begin{aligned} \mathrm{(1) \; \; \; \; \; \; \; \; \; v= u + at} \end{aligned}$$
$$\begin{aligned} \mathrm{(2) \; \; \; \; \; \; \; \; \; s= ut + \frac {1}{2}at^2} \end{aligned}$$
$$\begin{aligned} \mathrm{(3) \; \; \; \; \; \; \; \; \; v^2 = u^2 + 2as} \end{aligned}$$
Equations (1) and (3) are used to find final speeds . Eq. is the most simple and it is used when you know the time (how long the acceleration has been on) and equation 3 is used when you don´t know the time, but know the displacement during acceleration. Equation (2) is to find the displacement after the acceleration has been on for a known time t; the object may be already moving when the acceleration takes place. That is the meaning of the term ut in the equation. We need to know what equation to use , depending on the exercise or problem given.
Example calculation
Examples and solved problems (using the formulas above):
1) The acceleration of a car is 5 m / m/s2 . If it is initially at rest, what will be its velocity after 20 s?
Use formula (I):
u = 0 (initial speed)
v = at= 5*20 = 100 m/s
2) A car , initially travelling at 10 m/s, develops an acceleration of 6 m/s2 during 5 s. How far does it travel during this 5 s?
Use formula (II):
u=10m/s
a= 6m/s/s = 6 m/s2
s= distance travelled= 10*5+ 6* 5*5 / 2 = 125 m
3) Car A leaves from rest (u=0) , 10 miles down the road . It has an acceleration of 2 m/s2.
Car B leaves from the beginning of the road (u=0) and its initial speed is 10 m /s. Its acceleration is 5 m/s2.
How long will take to car B to meet car A? At what position the meeting will take place?
(Two cars travelling on the same road with different accelerations)
Use formula (II):
When the 2 cars meet, their position s must be the same. Firstly, let's write the expression for s in both cases:
car A: sa = 16000 + 2*t2/2
car B: sb= 10t+5*t2/2
Putting these 2 equations together we have an equation that can be solved to obtain t:
16000 + 2*t2/2 = 10t+5*t22
-3*t*t-10t+16000=0
3*t2+10t-16000=0
You just have to solve this second degree equation to get t. After that, you can plug t in any of the expressions for s (car A or B) in order to obtain the position where they meet.
4) An airport runaway has a length of 2 km. Considering that an airplane lands at one end of the runaway and manages to brake and come to rest at the other end, what is the maximum speed the plane can land at? Consider that the brakes provide an acceleration of -5 m/s2 .
Hint: on this problem we have a distance , an acceleration and a final speed, so that we must use formula III.