Question

X=2-4t+8t^2-t^3 Найти: V-? t-? a-? x-?​

Answer 1

скорость - производная х

ускорение - производная V

если исходное уравнение

x=2-4t+8t²-t³, то

V=-4+16t-3t²

a=16-6t

чтобы вычислить значения x V a  нужно подставить значение t

Answer 2

Given:

X = 2 - 4t + 8t^2 - t^3

To find:

V-? t-? a-? x-?​

We can use calculus to find the velocity, acceleration, and time at which the velocity is zero.

First, we find the derivative of X with respect to t to get the velocity function:

V = dX/dt = -4 + 16t - 3t^2

Next, we set the velocity function equal to zero to find the time(s) at which the velocity is zero:

-4 + 16t - 3t^2 = 0

Solving for t using the quadratic formula, we get:

t = (16 ± sqrt(256 - 4(-3)(-4))) / (2(-3))

t = (16 ± sqrt(112)) / (-6)

t ≈ 0.245, 2.205

These are the times at which the velocity is zero.

To find the acceleration at these times, we take the derivative of the velocity function:

a = dV/dt = 16 - 6t

Plugging in the times t ≈ 0.245 and t ≈ 2.205, we get:

a ≈ 14.32, -20.83

So at t ≈ 0.245, the acceleration is approximately 14.32, and at t ≈ 2.205, the acceleration is approximately -20.83.

Finally, to find the position and velocity at these times, we plug them into the original equations for X and V:

When t ≈ 0.245:

X ≈ 1.989

V ≈ 0

When t ≈ 2.205:

X ≈ -41.751

V ≈ 0

Therefore, the answers are:

V- 0 t- t ≈ 0.245, t ≈ 2.205 a- a ≈ 14.32, a ≈ -20.83 x- x ≈ 1.989 when t ≈ 0.245, x ≈ -41.751 when t ≈ 2.205