How Long Does It Take a Package to Land in a Moving Truck From an Overpass?

AI Thread Summary
The discussion centers on calculating the timing for a package dropped from an overpass to land in a moving truck below. The truck accelerates at 1/2g (4.9 m/s²) from rest, starting 100m away from the drop point, while the package falls 30m. The correct timing for the package drop is determined to be approximately 3.9 seconds after the truck starts moving. Participants express confusion over the equations needed to solve the problem, particularly regarding the initial velocities and accelerations involved. Ultimately, the solution involves calculating the time for both the truck and the package to reach the same point simultaneously.
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Homework Statement


A delivery truck wants to speed up its deliveries by dropping its packages into moving trucks. A worker is positioned in an overpass directly above a straight, level road to drop the packages into the truck at the correct time. One day, a delivery truck starts from rest and drives along the road with a constant acceleration of 1/2g. A package is released at the correct instant to land in the truck. If the overpass was 30m above the truck and the truck started from a position 100m from the point of impact, how long after the truck started did the employee wait before dropping the package? (Answer is 3.9s)

Homework Equations



The Attempt at a Solution


The givens I tried to draw out from this problem were:
Average acceleration = 1/2g = 1/2(9.8m/s^2) = 4.9m/s^2
Displacement of package being dropped to truck = 30m
Initial displacement of truck = 100m
Time = ?

I don't know what equations to use in order to solve this... I'm assuming we're dealing with 2 different things at once... but I just don't know what to do...
 
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You are trying to make the package and the truck arrive at the same point under the bridge at the same time. So ask yourself: How long does it take the truck to get there from it's starting point? How long does it take the package to get there from the bridge?
 
I'm still stumped... I tried using the equation:
d = vi(t) + Aav(t)^2/2
and substituted the givens of the package to find the time it took for the package to drop from 30m to 0m:

30m = 0(t) + 9.8m/s^2(t)^2/2
t = 2.47s

(I'm not sure if I'm supposed to be using 9.8 as the average acceleration...)
And then I don't exactly know what to do with that time as I'm still unsure about the givens of the truck...


Any hints on which equations I'm supposed to use...? And if I've gotten the right givens...?
 
fyesics said:

Homework Statement


A delivery truck wants to speed up its deliveries by dropping its packages into moving trucks. A worker is positioned in an overpass directly above a straight, level road to drop the packages into the truck at the correct time. One day, a delivery truck starts from rest and drives along the road with a constant acceleration of 1/2g. A package is released at the correct instant to land in the truck. If the overpass was 30m above the truck and the truck started from a position 100m from the point of impact, how long after the truck started did the employee wait before dropping the package? (Answer is 3.9s)

Homework Equations



The Attempt at a Solution


The givens I tried to draw out from this problem were:
Average acceleration = 1/2g = 1/2(9.8m/s^2) = 4.9m/s^2
Displacement of package being dropped to truck = 30m
Initial displacement of truck = 100m
Time = ?

I don't know what equations to use in order to solve this... I'm assuming we're dealing with 2 different things at once... but I just don't know what to do...
the truck takes √40 seconds to come and the packet takes √6 seconds to reach truck. subtraction of the two is 3.88s
u should take initial vel. of truck and packet to be 0.
 
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