Question

1. Уравнение движения материальной точки по прямой имеет вид x=A Bt Ct2 , где А=2м, В=2 м/с, С=-0,5 м/с2. Найти момент времени, в который скорость точки v=0. Чему равны координата x и ускорение точки а в этот момент времени?

Answer 1

1. Запишем уравнение: x=2+2·t-0,5·t²
Продифференцируем уравнение и получим скорость этой точки:
x'=v=2-t ⇒ при v=0 , то t=2 cекунды.
x(t)=2+2·2-0,5·4=4 м.
Чтобы найти ускорение в момент времени t, нужно дважды продифференцировать уравнение, то есть:
x''=a=-1 м/с²

Answer 2

I will answer your question in detail. 1. The equation of motion of a material point along a straight line has the form x=A+Bt+Ct^2, where A=2 m, B=2 m/s, C=-0.5 m/s^2. Find the moment of time when the speed of the point v=0. What are the coordinate x and the acceleration of the point a at this moment of time?

To find the moment of time when the speed of the point v=0, we need to use the formula for the velocity of the point, which is obtained by differentiating the equation of motion with respect to time:

v = dx/dt = B + 2Ct

To find the value of t when v=0, we equate the expression to zero and solve for t:

B + 2Ct = 0

t = -B/2C

Substituting the given values of B and C, we get:

t = -2/(-0.5) = 4 s

This means that the speed of the point becomes zero after 4 seconds of motion.

To find the coordinate x and the acceleration of the point a at this moment of time, we need to substitute the value of t into the equations of motion and acceleration, respectively:

x = A + Bt + Ct^2

a = dv/dt = 2C

Substituting the given values of A, B, C and t, we get:

x = 2 + 2*4 + (-0.5)*4^2 = 10 m

a = 2*(-0.5) = -1 m/s^2

This means that the point reaches the position of 10 meters from the origin and has a negative acceleration of 1 m/s^2 at the moment of time when its speed becomes zero.