Accelerated movement: variable velocity/ constant acceleration
When an acceleration is present, the velocity changes accordingly (formulae on the left side of the page).
If the acceleration is positive the velocity increases and when it is negative the velocity decreases (braking).
The acceleration of gravity, g, is often present in exam questions.
The units for acceleration are meters per second per second. This is the unit for velocity divided per time. It can also be described as meter per square second.
Examples:
1) A car is accelerated at 5 m/s/s. It is initially at rest. What will be its speed after 20 s?
v0 = 0 (initial speed)
so, v = at= 5*20 = 100 m/s
2) A car , initially travelling at 10 m/s, develop an acceleration of 6 m/s/s during 5 s. How far does it travel during this 5 s?
Now we will use the formula below.
v0=10m/s
a= 6m/s/s
s-s0 = distance travelled= 10*5+ 6* 5*5 / 2 = 125 m
3) Two cars travel in the same road with different accelerations.
Car A leaves from rest (vo=0) , 10 miles down the road .It has an acceleration of 2 m/s/s.
Car B leaves from the beginning of the road (s0=0) and its initial speed is 10 m /s. Its acceleration is 5 m/s/s.
How long will take to car B to meet car A? At what position the meeting will take place?
Let's apply the long formula again.
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: s = 16000 + 2*t*t/2
car B: s = 10t+5*t*t/2
Putting these 2 equations together we have an equation that can be solved to obtain t:
16000 + 2*t*t/2 = 10t+5*t*t/2
-3*t*t-10t+16000=0
3*t*t+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.
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