Question

Найдите V, если x(t)=t^4+t^3 +t^2+t ;t=2

Answer 1

X(t)=t⁴+t³+t²+t


V(t)=x'(t)=4t³+3t²+2t+1

V(2)=4*2³+3*2²+2*2+1=32+12+5=49

Answer 2

To find the velocity (V) at a specific time (t), you need to find the derivative of the position function (x(t)) with respect to time (t). The velocity is the first derivative of the position function.

Given the position function: x(t) = t^4 + t^3 + t^2 + t

To find the velocity (V), take the derivative of x(t) with respect to t: V = dx/dt

V = d/dt (t^4 + t^3 + t^2 + t)

Now, let's find the derivative step-by-step:

d/dt (t^4) = 4t^3 d/dt (t^3) = 3t^2 d/dt (t^2) = 2t d/dt (t) = 1

Now, combine all the derivatives:

V = 4t^3 + 3t^2 + 2t + 1

Now, we can find the velocity (V) at t = 2:

V = 4(2)^3 + 3(2)^2 + 2(2) + 1 V = 4(8) + 3(4) + 4 + 1 V = 32 + 12 + 5 V = 49

So, the velocity (V) at t = 2 is 49.