fisica1-youn-e-freedman-exercicios-resolvidos

3.91: In a time ∆ t, the velocity vector has moved through an angle (in radians)
(see Figure 3.23). By considering the isosceles triangle formed by the two velocity
vectors, the magnitude ∆ v r is seen to be 2v sin( φ / 2)
, so that
v∆t
∆φ =
R
v ⎛ v∆t
⎞ 10 m/s
a v
ave
= 2 sin⎜
⎟ = sin(1.0/s ⋅ ∆t)
∆t
⎝ 2R
⎠ ∆t
2
2
Using the given values gives magnitudes of 9 .59 m/s , 9.98 m/s and 10.0 m/s
2 . The
2
2
2
instantaneous acceleration magnitude, v / R = (5.00 m/s) /(2.50 m)
= 10.
0 m/s is indeed
approached in the limit at ∆t → 0.
The changes in direction of the velocity vectors are
v∆t
given by ∆θ = and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle
R
of the average acceleration vector with the original velocity vector is
π + ∆θ
= π / 2 + 1/ 2 rad(118.6°
), π / 2 + 0.1rad(95.7 ), and π / 2 + 0.05 rad(92.9°
).
2
°