The velocity as a function of time for an asteroid in the asteroid belt is given

Question

The velocity as a function of time for an asteroid in the asteroid belt is given by

where v0 and t0 are constants.

Use ti=0 as the initial time, and tf=1.4t0 as the final time.

The values for the constants that you will use are:

v0 = 3.5 m/s
t0 = 816 s

Find the displacement of the asteroid.
Find the x component of the acceleration at the final time.

Find the speed of the asteroid at the final time.

Find the y-component of the average velocity of the asteroid.

Find the x-component of the average velocity of the asteroid.

Find the magnitude of the acceleration at the final time.

Find the magnitude of the average acceleration of the asteroid.

Explanation / Answer

Here v(t) = v0e-t/t0 i + v0t/2t0 j

The x-component of velocity is vx(t) = v0e-t/t0 (the i component of the vector) and

y-component of velocity is vy(t) = v0t/2t0

The time derivative of a vector v is acceleration

ax = dvx(t)/dt = -v0/t0*e^-t/t0 evaluated at t = final time
If a is the acceleration vector, that is a = ax*i + ay*j
ax = -v0/t0*e-t/t0 = 3.5/(816*e^1.4) = 1.05*10^-3 m/s2

ay = dvy(t)/dt = v0/2t0 = 3.5/(2*816) = 2.144*10^-3 m/s2

Average Acceleration = sqrt(1.05^2 + 2.144^2)*10^-3 = 2.38*10^-3 m/s2

The position vector r(t) is v(t)dt = -v0*t0*e^-(t/t0)*i + [v0*t²/2]*j + C

r(0) = 0 at t = 0 then -v0*t0*i = C and C = v0*t0*i
r(t) = v0*t0*[1 - e^=(t/t0)]*i + [v0*t²/2]*j
dr = r(tf) - r(ti)
aavg = [v(tg) - v(ti)]/(tf - ti) (final velocity minus initial velocity divided by the time)
vavg =[s(tf) - s(ti)]/(tf - ti) (final position minus initial position divided by the time)
speed = |v| = sqrt[vx² + vy²]

vx = v0*e^(-t/t0) = 3.5*e^(-1.4) = 0.863 m/s and vy = [v0*t/(2t0)] = 3.5*1.4/2= 2.45 m/s

average speed = ((0.863)^2 + (2.45)^2)^0.5 = 2.597 m/s

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