The acceleration of a certain rocket is given by a x = bt , where b is a positiv

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 = 2.8 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 = 2.8 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

A)

Acceleration

ax = bt

Velocity

v(t) = axdt

v(t) = btdt

v(t) = bt2/2

Displacement

x(t) = vdt

x(t) = (bt2/2)dt

x(t) = (b/2)t2dt

x(t) = (b/2)(t3/3)

x(t) = bt3/6

B)

x0 = 0,

v0 = 0 and

b = 2.8 m/s3.
x(5.1) = bt3/6

x(5.1) = 2.8 x 5.13/6

x(5.1) = 61.9038 m
v(5.1 s) = bt2/2

v(5.1 s) =2.8 x 5.12/2

v(5.1 s) = 36.414 m/s

C)

x(4.6) = bt3/6

x(4.6) = 2.8 x 4.63/6

x(4.6) = 45.42 m

x(5.6) = bt3/6

x(5.6) = 2.8 x 5.63/6

x(5.6) = 81.95 m

Vavg = {x(5.6)- x(4.6)}/(5.6-4.6)

Vavg = (81.95-45.42)/1

vavg = 36.53 m/s

v(5.1 s) = 36.414 m/s

Yes, average velocity is in good agreement with the instantaneous velocity at t = 5.1 s

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