Question

A particle's position is given by x = 4.0t2 - 32t + 36, with x in meters and time t in seconds. At what position does the particle come to a momentary stop, to reverse its motion?

Answer 1

To find the stopping position of a particle, we first obtain the velocity by differentiating its position function, set the velocity to zero, and solve for time. Using the obtained time, we then calculate the position. For the position function x(t) = 4.0t^2 - 32t + 36, the particle stops momentarily at -28 meters.

Step-by-step explanation:

To find the position at which a particle comes to a momentary stop before reversing its motion, we need to determine when its velocity is zero. The velocity is the first derivative of the position function with respect to time. Given the position function x(t) = 4.0t^2 - 32t + 36, we can find the velocity by differentiating it to get v(t) = 8.0t - 32. Setting this velocity function to zero and solving for t, we have:

0 = 8.0t - 32
8.0t = 32
t = 4 seconds

Now, we can substitute t = 4 seconds back into the original position equation to determine the position at which the particle stops:

x(4) = 4.0(4)^2 - 32(4) + 36
= 64 - 128 + 36
= -28 meters

Therefore, the particle comes to a momentary stop at -28 meters from the origin.

Answer 2

firstly, we must find the equation the speed, it can be obtained by the derivative of x =x = 4.0t2 - 32t + 36, it means v= 8.0t-32. the particule stops means v= 0, and 0= 8.0t-32, which implies 8t=32, so t=4s, and then x(t=4) = 4.0(4)^2 - 32(4) + 36= - 28m
so x = - 28 m