Question

A snowmobile is originally at the point with position vector 31.7 m at 95.0° counterclockwise from the x axis, moving with velocity 4.69 m/s at 40.0°. It moves with constant acceleration 2.08 m/s2 at 200°. After 5.00 s have elapsed, find the following. (a) its velocity vector v? (b) its position vector r ?

Answer 1

(a) Here, x - component of required velocity vector = 4.69*cos 40 + 2.08*5*cos 200 = 3.59 - 9.77 = - 6.18 m/s
y component of required velocity vector = 4.69*sin 40 + 2.08*5*sin 200 = 3.01 - 3.56 = -0.55 m/s

So,
magnitude of vector = sq rt[(-6.18)^2 + (-0.55)^2] = 6.20 m/s
and theta = arctan[-0.55/ -6.18] = 185°

(b) Here, x component of required position vector = 31.7*cos 95 + 5*4.69*cos 40 + 0.5*25*2.08*cos 200
= -2.76 + 18.0 -24.43 = -9.19 m

y component of required position vector = 31.7*sin 95 + 5*4.69*sin 40 + 0.5*25*2.08*sin 200
= 31.58 + 15.07 - 8.90 = 37.75 m

magnitude of position vector = sq rt[(-9.19)^2 + (37.75)^2] = 38.85 m
and angle with x axis = arctan[37.75/(-9.19)] = 103.7°