Problem 2- A car is stopped at a traffic light. It then travels along a straight road so that its position from the light is given by x(t) 2.5t - 0.13t^2 in meters.
c. What is the position of the particle when it reverses its motion?
x=2.5t-1.3t^2
I assume you have some calculus, so max x is the derivative of x set to zero, so
0=2.5-2.6t solve for t, then put that in the original equation and solve for x.
Thank you
To find the position of the particle when it reverses its motion, we need to determine the point in time when the velocity of the car becomes zero.
The velocity of the car can be obtained by differentiating the position function with respect to time, t:
v(t) = d/dt(x(t)) = d/dt(2.5t - 0.13t^2) = 2.5 - 0.26t
Now, let's set v(t) = 0 and solve for t:
2.5 - 0.26t = 0
By solving this equation, we can determine the value of t when the velocity of the car becomes zero:
0.26t = 2.5
t = 2.5 / 0.26 ≈ 9.62 seconds
Now that we know the time at which the velocity is zero, we can substitute this value back into the position function to find the position when the car reverses its motion:
x(t) = 2.5t - 0.13t^2
x(9.62) = 2.5 * 9.62 - 0.13 * (9.62)^2
x(9.62) ≈ 24.05 - 11.89 ≈ 12.16 meters
Therefore, the position of the car when it reverses its motion is approximately 12.16 meters.