Answer: D. Acceleration of Object A is twice of that of the acceleration of Object B.
1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):
![v=\frac{S}{t}](https://tex.z-dn.net/?f=%20v%3D%5Cfrac%7BS%7D%7Bt%7D%20%20)
Rearranging the equation, we can write
![S=vt](https://tex.z-dn.net/?f=%20S%3Dvt%20)
In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is
![S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m](https://tex.z-dn.net/?f=%20S%3Dvt%3D%2814000%20m%2Fs%29%28150%20s%29%3D2.1%20%5Ccdot%2010%5E6%20m%20)
2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:
![S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s](https://tex.z-dn.net/?f=%20S%3Dvt%3D%2814000%20m%2Fs%29%28300%20s%29%3D4.2%20%5Ccdot%2010%5E6%20m%2Fs%20)
Hi there!
Distance and displacement are different.
Distance is WITHOUT direction (scalar quantity)
Displacement TAKES INTO ACCOUNT direction (vector quantity), and is RELATIVE to the initial point.
DISTANCE:
Add up each segment's position:
0 - 2 sec: 0 m
2 - 5 sec: 10 - 4 = 6 m
5 - 6 sec: 0 m
6 - 8 sec: 10 - 2 = 8 m
8 - 9 sec: 0 m
9 - 12 sec: 5 - 2 = 3 m
Add up:
6 + 8 + 3 = 17 m
DISPLACEMENT:
To find displacement, simply look at the position at the final point (t = 12 sec) and compare to the initial position:
At t = 12: Position = 5 m
At t = 0: Position = 4 m
5 - 4 = 1 m
The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.