Question

The equation for the motion of an object with constant acceleration is d=vt+0.5at2where d is distance traveled in ft, v is starting velocity in ft/s, a is acceleration inft/s2, and t is time in seconds. A racecar begins with a velocity of 200 ft/s andaccelerates at 100 ft/s2.Write an expression for the distance the car travels after t seconds.What is the GdF of the expression?

Answer 1

Given data:

The given initial velocity is v=200 ft/s.

The given acceleration is a=100 ft/s^2.

The expression for distance traveled by the body is,

$d=vt+\frac{1}{2}at^2$

Substitute the given values in the above expression.

$\begin{gathered} d=(200)(t)+\frac{1}{2}(100)t^2 \\ =200t+50t^2 \end{gathered}$

Thus, the expression for the distance covered at any time t is d=200t+50t^2.