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A bolt drops from the ceiling of a moving train car that is accelerating northward at a rate of 2.50 $\mathrm{m} / \mathrm{s}^{2}$ . (a) What is the acceleration of the bolt relative to the train car' (b) What is the acceleration of the bolt relative to the Earth? (c) Describe the trajectory of the bolt as seen by an observer fixed on the Earth.

(a) $\left| \vec { a } _ { \mathrm { BC } } \right| = 10.11 \mathrm { ms } ^ { - 2 }$

(b) $a _ { \mathrm { BE } } = 9.8 \mathrm { ms } ^ { - 2 }$

(c) The path of the bolt is a parabolic path.

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for this problem on the topic Off motion in two dimensions were told that a boat drops from the ceiling off a moving train car and were given the acceleration off the moving train car. We want to calculate the acceleration off the boat relative to the train car, the exploration of the same ball relative to earth as well as the trajectory of the bolt as seen by any fixed observer on Earth. Now, firstly will choose the reference frame with the positive X access to me, not with direction and the positive y axis vertically upward in our diagram. Now the acceleration of the car and the boat in bolt, which is in free fall related toe out A e, which is the acceleration off the car relative to earth. 2.5 m per square. Second in the positive X direction and a e. The acceleration of the alternative to Earth, which is 9.8 m per square second in the negative y direction. Now thes accelerations are related to the acceleration off the boat relative to the car by the following expressions. So a BC is equal to a the minus a see the exploration of the board, right toe minus exploration off the train car. Richard Toe. Or we can write this as a E is equal to a see minus or plus rather a B C so we can see that these relative accelerations from the 90 degree Victor Triangle has shown above. So first we want to find a B C. Now ABC has X and y components as follows The X component off ABC is equal to you can see from the vector diagram minus the X component of a c e. You can see that these two x components coincide and we know the X component of a C e. These 2.5. So this is minus 2.5 meters a square second, and the Y component of ABC corresponds to the Y component or minus the y component of a B. Okay. And we know this is 9.8 m per square second. So the Y component of ABC is minus 9.8 m per square second acceleration due to gravity. Me So therefore we have our vector a d. C. So the separation of the ball relative to the train car is to 0.5 m per square second southwards and nine 0.8 m per square second downwards. So towards the floor off the train. And that's our acceleration off the falling ball relative to the moving train car. That's all on the plot. Eight. Now for part B, we want to find the exploration of the boat relative to Earth. So they're moving the train. Come from this. On this scenario, we know that the acceleration of the ball relative to the earth is that simply of a free falling body. So this acceleration a E is simply nine 0.8 m per square second toward the earth. So in this case, this is downward. Yeah. Who finally You want to describe the trajectory off the boat have seen by any observer and you fix those up on up. Now, any observer fixed on art will see the bolt follow a parabolic park, and the parabolic part will be about in vertical access. And this is the same as any freely falling body having an initial horizontal velocity