Answer:
M a = (M1 + M2) a = F Newton's Second Law
F = (M2 - M1) g net force on the system
a = (M2 - M1) / (M1 + M2) g
a = (9 - 7) / (9 + 7) g = 2 / 16 * 10.0 m/s^2 = 1.25 m/s^2
Answer:
1 KM per minute is the real speed in minutes, turn that into 1000 meters per minute and divided by 60, you get a good number of 16.6666666667 which means you could go 50 meters per 3 seconds
Explanation:
so it would be 16.6666666667 meters per second
Answer:

Explanation:
a(t)=20t i+sin(t) j +cos(2t) k
v(t)=
=
---------------eqn 1
given v(0)=i
i=

from equation 1
V(t)=
----------eqn 2
now r(t)=

given r(0)=j
0i+1j+ok = 


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.