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# LINEAR MOTION PowerPoint PPT Presentation

LINEAR MOTION. TICKER TAPE.

LINEAR MOTION

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## TICKER TAPE

A ticker-timer consists of an electrical vibrator which vibrates 50 times per second. This enables it to make 50 dots per second on a ticker-tape being pulled through it. The time interval between two adjacent dots on the ticker-tape is called one tick. One tick is equal to 1/50 s or 0.02 s.

### UNIFORM VELOCITY

• The distance of the dots is equally distributed.

• All lengths of tape in the chart are of equal length.

• The object is moving at a uniform velocity.

### UNIFORM ACCELERATION

• The distance between the dots increases uniformly.

• The length of the strips of tape in the chart increase uniformly.

• The velocity of the object is increasing uniformly, i.e. the object is moving at a constant acceleration.

### UNIFORM DECELERATION

• The distance between the dots decreases uniformly.

• The length of the strips of tape in the chart decreases uniformly.

• The velocity of the object is decreasing uniformly, i.e. the object is decelerating uniformly.

## EXAMPLE 1

Diagram 2.4 shows a strip of ticker tape that was pulled through a ticker tape timer that vibrated at 50 times a second. What is thea. time taken from the first dot to the last dot?b. average velocity of the object that is represented by the ticker tape?

• a. There are 15 ticks from the first dot to the last dot, henceTime taken = 15 × 0.02s = 0.3sb. Distance travelled = 15cm

v=s/t

v=15cm/0.3s

v=50cms-1

## EXAMPLE 2

The ticker-tape in figure above was produced by a toy car moving down a tilted runway. If the ticker-tape timer produced 50 dots per second, find the acceleration of the toy car.

## EXAMPLE 3

A trolley is pushed up a slope. Diagram above shows ticker tape chart that show the movement of the trolley. Every section of the tape contains 5 ticks. If the ticker-tape timer produced 50 dots per second, determine the acceleration of the trolley.

### EXAMPLE 1

• A car is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and the driver applies the brakes and skids to a stop. If the car acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process.

• Given:

vi = +30.0 m/s vf = 0 m/s

a = - 8.00 m/s2

• (0 m/s)2 = (30.0 m/s)2 + 2(-8.00 m/s2)s

0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)s

(16.0 m/s2)s = 900 m2/s2 - 0 m2/s2

(16.0 m/s2)s = 900 m2/s2

s = (900 m2/s2)/ (16.0 m/s2)

s = (900 m2/s2)/ (16.0 m/s2)

s = 56.3 m

### EXAMPLE 2

• A bus is waiting at a stoplight. When it finally turns green, the bus accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine displacement of the bus during this time period.

• Given:

vi = 0 m/s t = 4.10 s

a = 6.00 m/s2

• s = (0 m/s)(4.1 s) + 0.5(6.00 m/s2)*(4.10 s)2

s = (0 m) + 0.5(6.00 m/s2)(16.81 s2)

s = 0 m + 50.43 m

s = 50.4 m

### PROBLEM 1

• An airplane accelerates down a run-way at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before take-off.

• d = vi*t + 0.5*a*t2

• d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

• d = 1720 m

### PROBLEM 2

• A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

• d = vi*t + 0.5*a*t2

• 110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2

110 m = (13.57 s2)*a

a = (110 m)/(13.57 s2)

a = 8.10 m/ s2

### PROBLEM 3

• It was once recorded that a Jaguar left skid marks which were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.