suppose that the position of a particle as a function of time is given as:
x(t)=1t^4(i^hat)+(-2t^3+1t^4)j^hat
a) determine velocity as a function of time:
b)determine acceleration as a function of time:
c)determine the magnitude of the velocity at t=0.7s
a) To determine the velocity as a function of time, we need to differentiate the position function x(t) with respect to time.
Given x(t) = t^4 * i^hat + (-2t^3 + t^4) * j^hat
Differentiating with respect to time, we get:
v(t) = (d/dt)(t^4 * i^hat) + (d/dt)(-2t^3 + t^4) * j^hat
Since i^hat is a unit vector in the x-direction and j^hat is a unit vector in the y-direction, we can ignore them while differentiating.
So, v(t) = (d/dt)(t^4) * i^hat + (d/dt)(-2t^3 + t^4) * j^hat
Differentiating t^4 gives us:
v(t) = 4t^3 * i^hat + (-6t^2 + 4t^3) * j^hat
Thus, the velocity as a function of time is:
v(t) = 4t^3 * i^hat + (-6t^2 + 4t^3) * j^hat
b) To determine the acceleration as a function of time, we differentiate the velocity function v(t) with respect to time.
Given v(t) = 4t^3 * i^hat + (-6t^2 + 4t^3) * j^hat
Differentiating with respect to time, we get:
a(t) = (d/dt)(4t^3 * i^hat) + (d/dt)(-6t^2 + 4t^3) * j^hat
Again, i^hat and j^hat can be ignored while differentiating.
So, a(t) = (d/dt)(4t^3) * i^hat + (d/dt)(-6t^2 + 4t^3) * j^hat
Differentiating 4t^3 gives us:
a(t) = 12t^2 * i^hat + (-12t) * j^hat
Thus, the acceleration as a function of time is:
a(t) = 12t^2 * i^hat + (-12t) * j^hat
c) To determine the magnitude of the velocity at t = 0.7s, we substitute t = 0.7 into the velocity function v(t) and find its magnitude.
Given v(t) = 4t^3 * i^hat + (-6t^2 + 4t^3) * j^hat
Substituting t = 0.7, we get:
v(0.7) = 4(0.7)^3 * i^hat + (-6(0.7)^2 + 4(0.7)^3) * j^hat
Calculating the values, we get:
v(0.7) = 0.0196 * i^hat + 0.4116 * j^hat
The magnitude of the velocity is given by:
|v(0.7)| = sqrt((0.0196)^2 + (0.4116)^2)
Evaluating this expression, we find:
|v(0.7)| ≈ 0.412 m/s
To determine the velocity and acceleration as functions of time, we need to find the derivatives of the position function with respect to time.
a) Velocity as a function of time:
The velocity of a particle is the derivative of its position function with respect to time. In this case, we have:
x(t) = t^4 i^hat + (-2t^3 + t^4) j^hat
To find the velocity, we take the derivative of x(t) with respect to time:
v(t) = d/dt (x(t))
= d/dt (t^4 i^hat) + d/dt (-2t^3 + t^4) j^hat
Taking the derivative with respect to time, we get:
v(t) = 4t^3 i^hat + (-6t^2 + 4t^3) j^hat
So, the velocity as a function of time is given by:
v(t) = 4t^3 i^hat + (-6t^2 + 4t^3) j^hat
b) Acceleration as a function of time:
Similarly, to find the acceleration as a function of time, we take the derivative of the velocity function with respect to time:
a(t) = d/dt (v(t))
= d/dt (4t^3 i^hat) + d/dt (-6t^2 + 4t^3) j^hat
Taking the derivative with respect to time, we get:
a(t) = 12t^2 i^hat + (-12t + 12t^2) j^hat
So, the acceleration as a function of time is given by:
a(t) = 12t^2 i^hat + (-12t + 12t^2) j^hat
c) Magnitude of velocity at t = 0.7s:
To determine the magnitude of the velocity at t = 0.7s, we substitute t = 0.7s into the velocity function we found in part (a):
v(0.7) = 4(0.7)^3 i^hat + (-6(0.7)^2 + 4(0.7)^3) j^hat
Calculating the values, we get:
v(0.7) ≈ 1.225 i^hat - 1.041 j^hat
To find the magnitude, we use the Pythagorean theorem:
|v(0.7)| = sqrt((1.225)^2 + (-1.041)^2)
Calculating the magnitude, we have:
|v(0.7)| ≈ sqrt(1.500 + 1.083)
Therefore, the magnitude of the velocity at t = 0.7s is approximately:
|v(0.7)| ≈ sqrt(2.583) ≈ 1.61
x = t^4 i + (-2t^3+1t^4) j ????
if so
dx/dx = 4 t^3 i + (-6 t^2 + 4 t^3) j = velocity
d^2x/dt^2 = 12 t^2 i +(-12 t + 12t^2) j = 12 [ t^2 i + (t^2-t) j = acceleration
at t = 0.7
dx/dt = (4 * 0.7^3) i + (4 * 0.7^3 - 6 *0.7^2) j
= 1.372 i + (1.372 - 2.94) j
= 1.372 i - 1.57 j
|v| = sqrt (1.37^2 + 1.57^2)