A student waiting at a stoplight notices that her turn signal, which has a period of 0.80 s, makes one blink exactly in sync with the turn signal of the car in front of her. The blinker of the car ahead then starts to get ahead, but 15 s later the two are exactly in sync again. Part A What is the period of the blinker of the other car

he period of the blinker of the other car = 0.747 s

Explanation:

The period of the blinker of the 1st car = 0.80 seconds

The other came is 1/15 times slower than the first car = 1/15 * (0.80) = 0.053

The period of the the blinker of the other car = Period of the blinker of the first car - (Difference in the periods of the blinkers of the two cars)

The period of the blinker of the other car = 0.80 - 0.053

The period of the blinker of the other car = 0.747 s

0.7593 seconds

Explanation:

A student waiting at a stoplight notices that her turn signal, which has a period of 0.80 s, makes one blink exactly in sync with the turn signal of the car in front of her. The blinker of the car ahead then starts to get ahead, but 14 s later the two are exactly in sync again. What is the period of the blinker of the other car?

Speed:

If a car is moving on a track and car covers x meter distance in y second time. Now the car driver want to know the distance covered by the car in unit time. So the distance covered in unit time will be calculated by dividing x meters by y seconds. This quantity is known as the speed.

Given:

The period of the blinker of the student car is t1 = 0.80

The period after which both blinks sync is t2 = 15seconds

Let the period of the blinker of the other car be t2

The blinker of the student car is 1 second and can be calculated as:

Student car blink / sec = 1/0.80

The blink of the other car is in 1 second and can be calculated as:

Other car blink/sec = 1/t2

According to the condition,

Other car blink/sec - students car blink / sec = 1/15

1/t2 - 1/0.80 = 1/15

1/t2 = 1/15 - 1/0.80

t2 = 0.7593

t2 = 0.7593second