Answer: The correct answer is option (A).
Explanation:
Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:



Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.
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:
57.1 km/hr
Explanation:
To find the average speed you take the total distance divided by the total elapsed time.
So, the total distance is 140 + 60 = 200
the total elapsed time is found by taking 140/70=2 and 60/40=1.5
2+1.5=3.5
The plug the numbers into the equation,
200/3.5=57.1