Question #17480

Q2. The altitude of a rocket in the first half-minute of its ascent is given by x = bt2, where x is the altitude and b = 2.90 m/s2 is a constant.
a) Find a general expression for the rocket’s velocity as a function of time.
b) Find the rocket’s instantaneous velocity at t = 20 s.
c) Find an expression for the average velocity of the rocket. Find the average velocity after during the first 20 seconds.
d) Compare your two velocities (instantaneous and average). Is there a difference and if so why?

Expert's answer

Q2. The altitude of a rocket in the first half-minute of its ascent is given by x=bt2x = bt2, where xx is the altitude and b=2.90m/s2b = 2.90 \, \text{m/s}^2 is a constant.

a) Find a general expression for the rocket's velocity as a function of time.

b) Find the rocket's instantaneous velocity at t=20st = 20 \, \text{s}.

c) Find an expression for the average velocity of the rocket. Find the average velocity after during the first 20 seconds.

d) Compare your two velocities (instantaneous and average). Is there a difference and if so why?

Solution:

Such as v=dXdtv = \frac{dX}{dt}

v=(bt2)=2btv = (bt^2)' = 2btv=2.92t=5.8tv = 2.9 * 2t = 5.8tv(t=20s)=5.820=116m/sv(t = 20 \, \text{s}) = 5.8 * 20 = 116 \, \text{m/s}v(average)=v0+vt2v(\text{average}) = \frac{v_0 + v_t}{2}v(averaget=20)=v0+v(t=20)2=0+1162=58m/sv(\text{average} \, t = 20) = \frac{v_0 + v(t = 20)}{2} = \frac{0 + 116}{2} = 58 \, \text{m/s}


Answers: v=5.8tv = 5.8t, v=116m/sv = 116 \, \text{m/s}, v(average)=v0+vt2v(\text{average}) = \frac{v_0 + v_t}{2}, v(averaget=20)=58m/sv(\text{average} \, t = 20) = 58 \, \text{m/s}

The instantaneous and average velocities are different.

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