You are in the back of a pickup truck on a warm summer day and you have just finished eating an apple. The core is in your hand and you notice the truck is just passing an open dumpster 7.0 m due west of you. The truck is going 30.0 km/h due north - cont?

Question

and you can throw that core at 60.0 km/h. In what direction should you throw it to put it in the dumpster, and how long will it take it to reach its destination?

Elena Albarran · Accepted Answer

My vantage point in the truck:
#v(t) ~~ 60j - 10*7/10k = 60j - 7k # I am rounding #g -> 10#
#time, t = 7/10 s#

#v(t) =v_(x)i +v_yj - "gt"k#
#v_(x)hatx+v_yhaty - "gt"hatz = ((v_x),(v_y),("-gt")) = ((-30),(60),("-9.81t"))# or

4) #v(t) = -30i + 60j - 7k#
The direction is given in the x-y plane is give by the angle between
the vector given by #(-30i + 60j); theta = tan^-1(-2) = -63.4^0# or #296.5^0#

Remark: You can also use conservation of momentum to get the direction. I have added the z direction because the core will influenced by gravity, thus will undergo a parabolic motion as it travels to dumpster...

Observer outside the truck vantage point

This is a great question that illustrate relative displacement and velocity, or in general acceleration. While your question does not touch on it the general consideration of this is to determine the ball trajectory in the presence #v_y, -v_x " and " a_z = g#. I will try to give you insight into both the simplified 2-D and 3-D views of the problem. I will do this from my reference point in the truck (which is what your question asks) and from an observer outside the train. Observer - Inside the truck, Me : The core will move with the constant velocity, #v_"North" = v_y = 60 m/s# away from the train. There is nothing that slows the core. So I will see the ball right in front of me, flying further away and falling down with #v_z=−gt# obviously, there will be a curved trajectory, a parabola in the y-z, the plane where the train is moving perpendicularly to. So what I see is the vector, 1) #v(t) =v_yj - "gt"k = v_yhaty - "gt"hatz = ((0),(v_y),("-gt")) = ((0),(v_y),("-9.81t"))# or 2) #v(t) = 60j - 9.81tk # To calculate t, you use the #v_y# and the distance to the dumpster distance #y = 7 m# #t = (7 m)/(60 m/s) = 7/60 s ~~.1167 # insert this in 2 and we have: 3) #v(t) ~~ 60j - 10*7/10k = 60j - 7k # I am rounding #g -> 10# Observer - Outside the truck, You clearly an observers on the side walk close to the truck will also see the speed of the truck so we need to adjust equation 1) and 2) as: 3) #v(t) =v_(x)i +v_yj - "gt"k# #v_(x)hatx+v_yhaty - "gt"hatz = ((v_x),(v_y),("-gt")) = ((-30),(60),("-9.81t"))# or 4) #v(t) = -30i + 60j - 7k # The direction is given in the x-y plane is give by the angle between the vector given by #(-30i + 60j); theta = tan^-1(-2) = -63.4^0# or #296.5^0#

Carson Alexander · Answer

undefined The apple core will fall with a horizontal velocity of 30.0 km/h due north and will also experience a downward acceleration due to gravity (9.8 m/s^2). Since the truck is moving at a constant velocity, there is no horizontal acceleration acting on the core. Thus, the horizontal motion of the apple core will remain constant at 30.0 km/h due north.

HIX Tutor · Answer

undefined Great start! Let's break this down step-by-step.

1. **Understand the scene**: You are in a truck moving north. The dumpster is to the west of you.

2. **Identify the distance**: The dumpster is 7.0 m away.

3. **Differentiating directions**: 
   - North is up on a map.
   - West is to the left on a map.

4. **Speed of the truck**: The truck is moving at 30.0 km/h. Let's convert that to meters per second (m/s) because we want to compare things in the same unit.
   - 30.0 km/h = 30,000 meters in one hour.
   - There are 3,600 seconds in an hour. So, 
   $$
   30,000 \text{ m} \div 3,600 \text{ s} \approx 8.33 \text{ m/s}.
   $$

5. **What are you thinking?**: If you want to throw the apple core into the dumpster, you need to consider:
   - The distance to the dumpster: 7.0 m west.
   - The speed of the truck moving north.

6. **Timing**: How long will it take for the truck to reach a certain point? For example, if the truck continues, how far will it go in a few seconds?

- If you want to throw the core while the truck is moving, you must think about how far the truck will go while you throw it.

7. **Next steps**: Do you want to calculate how far the truck gets after a few seconds? Or do you have a specific question about this situation?