Nonuniform linear motion
Definition of nonuniform linear motion
Nonuniform linear motion is motion at constant acceleration. In other words, nonuniform linear motion = motion with the magnification of acceleration is constant and the direction of acceleration is constant. Direction of acceleration is constant = direction of velocity is constant = direction of displacement is constant = direction of motion is constant = the object moves in straight line. The magnitude of constant acceleration means that the magnitude of velocity or speed increases regularly.
An object is initially at rest. One second later the object moves at a speed of 2 m/s. Two seconds later, the object moves at a speed of 4 m/s. Three seconds later, the object moves at a speed of 6 m/s. Four seconds later, the object moves at a speed of 8 m/s. And so on…
Every 1 second, the speed of the object increases by 2 m/s. The object experience a constant acceleration of 2 m/s per 1 second or 2 m/s per second or 2 m/s2.
The second example:
An object initially moves at a speed of 10 m/s. One second later
the speed is decreased to 9 m/s. Two seconds later, the speed decreases to 8 m/s. Three seconds later, the speed is decreased to 7 m/s. Four seconds later, the speed is decreased to 6 m/s. And so on … Every 1 second the speed of the object decreases by 1 m/s. The object has a constant deceleration of 1 m/s per 1 second or 1 m/s per second or 1 m/s2.
Equation of nonuniform linear motion
There are three equation formulas often used:
vt = vo + a t
d = vo t + 1⁄2 a t2
vt2 = vo2 + 2 a d
vo = initial speed (meter/second), vt = final speed (meter/second), a = acceleration (meter/second square ), t = time interval (second), d = distance (meter).
Sample problem 1:
A particle initially at rest, then experiences a constant acceleration of 2 m/s2 for 10 seconds. Calculate the final velocity of the particle.
Acceleration (a) = 2 m/s2 means speed (v) increase by 2 m/s each 1 second. After 2 second, v = 4 m/s. After 10 seconds, v = 20 m/s.
Known: vo = 0 m/s (particle at rest), a = 2 m/s2, t = 10 s
vt = vo + a t = 0 + (2)(10) = 20 m/s
Sample problem 2:
A car moves on a straight track at a speed of 60 km/h. Because there is an obstacle, the driver brakes so the car has a deceleration of 10 m/s2. What distance does the car still travel until it stops after braking is done?
vo = 60 km/jam = 60(1000 m)/3600 s = 17 m/s
vt = 0 m/s (car stop)
2a = – 10 m/s
Wanted: distance (d)
vt2 = vo2 + 2 a d
(0)2 = (17)2 + 2(-10) d
0 = 289 – 20 d
289 = 20 d
d = 289 : 20 = 14.45 meters
Sample problem 3:
The graph below is a nonuniform linear motion graph, v represents speed, and t states time. The magnitude of the acceleration of objects based on the graph is …
Sample problem 4:
The motion of a car produces a chart of speed (v) vs time (t) as shown in the figure on the side. If the area under the graph (shaded area) is 48 m, then the acceleration of the car is …
Shaded area = total distance = 48 meters
vo = 8 m/s, vt = 16 m/s, t = 1 m/s
vt = vo + a t
16 = 8 + a (1)
a = 16 – 8
a = 8 m/s2