AMA is actual mechanical advantage in where it would compare the force in as well as out in using actual measurements in the real life or with friction while in IMA which is the ideal mechanical advantage no friction then is needed. The answer then is a. the value of IMA is larger than AMA.
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.
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
A = (Vf-Vi) / t,
a = (6-0)/3 = 2m/s^2,
F = ma = 2 * 2 = 4N