Answer:
Explanation:
If we let our reference frame travel at 30 m/s with the constant speed car, The accelerating car increases its velocity by 10 m/s in 3 seconds.
The average velocity of the accelerating car is (0 + 10) / 2 = 5 m/s
It will advance its position 5 m/s(3 s) = 15 m in the accelerating period.
It takes 5 + 3 = 8 m for the two cars to become side by side.
It would take another 5 + 3 = 8 m for the accelerating car to leave a gap of 3 m between.
The car requires 8 + 8 = 16 m to pass the other safely but the acceleration period only gets him to 15 m.
So despite your saying this is not a YES / NO question, the answer is NO the acceleration is too low or not long enough to meet the required clearances.
Input needed is 10000 J/s / 0.30 = 333333 = J/s
three hours requires 333333(3)(3600) = 360 MJ of energy
360 MJ / 34 MJ/liter = 10.6 liters.
OK. So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.
-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.
-- You're pushing it with 5.0N of force.
-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)
-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.
The force on it to make it accelerate is (mass) x (acceleration) =
(7.4 kg) x (5/12.6 m/s²) = <em>2.936 N.</em>
The only force on the big box comes from the small box, pushing it from behind.
So that same <em>2.936N</em> must be the contact force between the boxes.
<span>The average speed is 14 m / sec or 50.4 km / h.</span>
We have to add two vectors.
Vector #1: 0.15 m/s north
Vector #2: 1.50 m/s east
Their sum:
Magnitude: √(0.15² + 1.50²)
Magnitude = √(0.0225+2.25)
Magnitude = √2.2725
Magnitude = <em>1.5075 m/s</em>
Direction = arctan(0.15/1.50) north of east
Direction = <em>5.71° north of east</em>
