Question

A race car driver is doing a time trial of a new car on a straight track. The car's distance from the timekeeper is represented by y=293x+50, where x is time in seconds and y is distance in feet from the timekeeper's position. How many feet from the timekeeper is the race car at the beginning? It is ft from the timekeeper.

Answer 1

Answer:

The distance at which the timekeeper is the race car at the start is 50 feet.

Step-by-step explanation:

You know that the car's distance from the timekeeper is represented by

y=293*x +50

where x is time in seconds and y is distance in feet from the timekeeper's position.

You want to determine how far the timekeeper is from the race car at the start. That is, the distance the timekeeper is from the car when the time is equal to zero.  This indicates that x = 0. Replacing x by that value in the expression of the distance of the car from the timekeeper as a function of time and solving:

y=293*0 +50

you get:

y=50

The distance at which the timekeeper is the race car at the start is 50 feet.

Answer 2

The distance from the timekeeper to the race car at the beginning is 50 feet.

• Let the time be x.

• Let the distance be y.

Given the following data:

• $y=293x+50$

• Time = 0 seconds (at the beginning).

To calculate how many feet (distance) from the timekeeper is the race car at the beginning:

You should note that, the time of the race at the beginning would be equal to zero (0) seconds.

Hence, the start time is equal to zero (0) seconds.

Substituting the value of x into the eqn, we have:

$y=293(0)+50\\\\y=50$

Distance = 50 feet.

Read more: https://brainly.com/question/14703849