The expression for the distance the car travels after t seconds is d = 200t + 50t²
<h3>Equation of motion </h3>
From the question, we are to write an expression for the distance the car travels after t seconds
From the given information,
The equation for the motion of an object with constant acceleration is
d = vt + 0.5at²
For the race car,
v = 200 ft/s
a = 100 ft/s²
Putting the parameters into the equation, we get
d = 200t + 0.5(100)t²
d = 200t + 0.5 ×100)t²
d = 200t + 50t²
Hence, the expression for the distance the car travels after t seconds is d = 200t + 50t²
Learn more on Equations of motion here: brainly.com/question/20910641
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you add the amount of people in each section that fits into the category. So, 11 people from 6-6:29 and 15 people from 6:30-6:59 and then 8 people from 7:30-7:59 so you get 34 people total
5.10508806
hope this helped!
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Answer:
I believe it would be 73.8 or if they are rounding up 74.
Step-by-step explanation:
15% of 60 is *9* and 8% of 60 is *4.8*
9+4.8=13.8
13.8+60=73.8.
