the position of an object is given by the position function x = 5.0t ^ 3 + 32t +
ID: 1886072 • Letter: T
Question
the position of an object is given by the position function x = 5.0t ^ 3 + 32t + 4.0, where x is in meters and t in seconds. Determine the following.A-the displacement between t1 = 1.0s & t2 = 3.0s
B-the average speed between t1 = 1.0s & t2 = 3.0s
C-the instantaneous velocity at t = 1.0s
D-average acceleration between t1 = 1.0s & t2 = 3.0s
E-instantaneous acceleration for t = 3.0s the position of an object is given by the position function x = 5.0t ^ 3 + 32t + 4.0, where x is in meters and t in seconds. Determine the following.
A-the displacement between t1 = 1.0s & t2 = 3.0s
B-the average speed between t1 = 1.0s & t2 = 3.0s
C-the instantaneous velocity at t = 1.0s
D-average acceleration between t1 = 1.0s & t2 = 3.0s
E-instantaneous acceleration for t = 3.0s
A-the displacement between t1 = 1.0s & t2 = 3.0s
B-the average speed between t1 = 1.0s & t2 = 3.0s
C-the instantaneous velocity at t = 1.0s
D-average acceleration between t1 = 1.0s & t2 = 3.0s
E-instantaneous acceleration for t = 3.0s
Explanation / Answer
given
x = 5.0*t^3 + 32*t + 4.0
A)
at t1 = 1.0s,
x1 = 5*1^3 + 32*1 + 4
= 41 m
at t2 = 3.0s,
x1 = 5*3^3 + 32*3 + 4
= 235 m
the displacement between t1 = 1.0s & t2 = 3.0s, x2 - x1 = 235 - 41
= 194 m
B)
The average speed between t1 = 1.0s & t2 = 3.0s, Vavg = (x2 - x1)/(t2 - t1)
= (235 - 41)/(3 - 1)
= 97 m/s
C) Instantaneous velocity, v = dx/dt
= 3*3*t^2 + 32 + 0
= 9*t^2 + 32
at t = 1 s,
v = 9*1^2 + 32
= 41 m/s
D) at t1 = 1s,
v1 = 9*1^2 + 32
= 41 m/s
at t2 = 3 s
v2 = 9*3^2 + 32
= 113 m/s
average acceleration between t1 = 1.0s & t2 = 3.0s, avg = (v2 - v1)/(t2 - t1)
= (113 - 41)/(3 - 1)
= 36 m/s^2
E) instantaneous acceleration, a = dv/dt
= 9*2*t + 0
= 18*t
at t = 3 s
a = 18*3
= 54 m/s^2
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