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

[fyesics]

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...

 

[Nugatory]

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?

 

[fyesics]

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...?

 

[altamashghazi]

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.

 

[asteriatic]

To solve this problem, we can use the equations of motion for uniform acceleration. The equations are:
1. v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time)
2. s = ut + 1/2at^2 (where s is displacement)
3. v^2 = u^2 + 2as

First, we need to find the final velocity of the truck when the package is dropped. Using equation 1, we have:
v = u + at
v = 0 + (1/2)(9.8)(t)
v = 4.9t m/s

Next, we can use equation 2 to find the displacement of the truck when the package is dropped. We know that the displacement of the package is 30m, so we have:
s = ut + 1/2at^2
30 = (0)t + 1/2(4.9)(t)^2
30 = 2.45t^2
t^2 = 30/2.45
t = √(30/2.45)
t = 3.9s

Therefore, the employee waited 3.9 seconds before dropping the package.