To solve letter a:
d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.
d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.
t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.
To solve letter b:
15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,
d = 115 * 0.71 = 81.65 km.
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Answer:
She will make the jump.
Explanation:
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
First we will consider horizontal motion of stunt women
Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0
Substituting

So she will cover 77 m in 2.85 seconds
Now considering vertical motion, up direction as positive
Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8
, time = 2.85
Substituting

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.
So she will make the jump.
Answer:
Shawn's speed relative to Susan's speed = 10 mph
Resultant velocity = 82.32 mph
Explanation:
The given data :-
i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.
ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.
iii) The speed of both Susan and Shawn is relative to earth.
iv) The angle between Susan in north and Shawn in east is 90°.
We have to find Shawn's speed relative to Susan's speed.
v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph
Resultant velocity,

v = 82.32 mph
Sound waves are longitudal waves meaning they go back and forth