Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.
It’s D. An enlargement (hope this helps!)
Answer:
Speed Unchanged.
Explanation:
As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.
Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.