The acceleration of a certain rocket is given by ax = bt, where b is a positive

Question

The acceleration of a certain rocket is given by ax = bt, where b is a positive constant. (a) Find the position function x(t) if x = x0 and v0 at t = 0. (Use the following as necessary: x0, v0, b, and t.) x(t) = (b) Find the position and velocity at t = 5.1 s if x0 = 0, v0 = 0 and b = 4.4 m/s3. x(5.1 s) = m v(5.1 s) = m/s (c) Compute the average velocity of the rocket between t = 4.6 s and 5.6 s at t = 5.1 s if x0 = 0, v0 = 0 and b = 4.4 m/s3. vavg = m/s Is this average velocity in good agreement with the instantaneous velocity at t = 5.1 s? Yes No

Explanation / Answer

ax= bt .

dv/dt= bt

v=(1/2)bt2+v0 .

and similarly dx/dt=bt2+v0 .

a)so x(t)= bt3/6 +vot +xo .

b)if xo=0 and vo=0 .

x(t)= bt3/6 and v(t)=(1/2)bt2 .

x(5.1)= 122 m and v(5.1) = 66 .5 m/s .

c)average velocity is (vdt)/dt .

so taking integration from 4.6 to 5.6 .

we get [bt3/6]5.64.6 .

average velocity = 57.4 m/s .

instantaneous velocity at t= 5.1 is 66.5 m/s .

answer:NO , THEY ARE NOT EQUAL .

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