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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Answer:
16
Step-by-step explanation:
All you have to do 4/9 of 36 which is 16.
Hope this helps!!
It would be 55.97
To find this you find what 20% of 69.96 is and subtract that number from 69.96 (in this case that number would be 13.99) So you would find your answer of 55.97
(0,-3)
Was the two supposed to be on the top
