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:
B) In Quadrant IV
Step-by-step explanation:
You move 7 units to the right on the x-axis. Then you move 10 units down the y-axis. The point will then land on quadrant IV.
Answer:
48
Step-by-step explanation:
multiply 8x6 which is 48 .....8,16,24,32,40,48
The sum of the interior angles is 180°
