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If we are cycling, which would often come to a decrease or an uphill road. When we see a declining path, bicycle run by itself without having to pedal or brake. Eventually, it will increase the speed of the bike quickly. If we know there are free falling coconuts from the tree. Both are examples of application of uniformly accelerated motion in everyday life.
To investigate the presence of uniformly accelerated motion the same when we investigate the existence of uniform rectilinear motion, using the ticker timer. Mathematically acceleration produced by objects which have a uniformly accelerated motion or uniformly accelerated motion can be formulated as follows:
a = Δv / t = (vt – v0) / t
with a = acceleration (m/s2)
Δv = change in initial and final velocity (m/s)
vt = final velocity (m/s)
vo = initial velocity (m/s)
t = time (s)
In uniformly accelerated motion there is agreement that the sign is used to simplify the problem in analyzing the results of a uniformly accelerated motion of the material produces an acceleration (+) or acceleration (-), as follows:
- If the acceleration (+) is accelerated uniformly accelerated motion direction of the velocity or acceleration
- If the acceleration (-) is uniformly accelerated motion is slowed down or accelerated in contrast to the
speed
Examples of application of uniformly accelerated motion of matter as a material with a simple calculation exercises that can help us to practice:
A child with the use of roller skates glide on a declining path, with initial velocity 20 m/s. After 10 seconds, its speed to 35 m / s. Calculate the acceleration!Completion:
Note:
vo = 20 m/s
vt = 35 m/s
t = 10 s
Asked: a ?
Answer: a = (vt – v0) / t = (35 – 20) / (10) = 15 / 10 = 1,5 m/s^2
Thus, the child slid to the acceleration of 1,5 m/s^2
A motorcycle that is moving at a speed of 40 m / s braking with deceleration 2 m/s^2. How long the bike is moving up the speed to 10 m / s and up to stop?
Completion:
a. The time required to reach speeds of 10 m/s
Note:
vo = 40 m/s
vt = 10 m/s
a = - 2 m/s2
then:
a = (vt – v0) / t
-2 m/s^2 = (10 – 40) / t
t = (- 30) / (- 2)
t = 15 s
Thus, the time required to reach speeds of 10 m /s is a 15 second
b. The time it takes to stop
Note: vo = 40 m/s
vt = 0 m/s
a = - 2 m/s2
then :
a = (vt – v0) / t
-2 m/s^2 = (0 – 40) / t
t = (- 40) / (- 2)
t = 20 s
Thus, the time required of the motorbike to stop since conducted the braking is 20 seconds.
What will happen after the above exercises, the brain is so dilute it? We multiply train yourself to do physics problems, then we will better understand the concepts being studied.
1 Response to Uniform Accelerated Motion
It's a nice approach to uniform acceleration. I really like it. Thanks for sharing it.
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