Apickup truck and a hatchback car start at the same position. if the truck is moving at a constant 33.2m/s and the hatchback car starts from rest and accelerates at 5m/s/s, how far away do the cars meet up again?

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

= 33.2 m/s, a = 0 (since the velocity is constant), = 0

Using

s = 33.2t .......... eq (1)

Hatchback:

, = 0 m/s (since initial velocity is zero), = 0

Using

putting in the data we will get

now putting 's' value from eq (1)

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

Explanation:

initial speed of the pickup truck (Up) = 33.2 m/s

acceleration of the pickup truck (ap) = 0

initial speed of the hatchback = 0

acceleration of the hatchback (ah) = 5 m/s^{2}

how far away (s) do the cars meet up again?

from the equations of motion distance covered (s) = ut +

distance covered by the pickup = ut +

where

distance covered by the hatchback = ut +

where

        =  ......equation 2

when the cars meet, they both would have covered the same distance, therefore

now that we have the time it takes for both cars to meet, we can put the value of the time into equation 1 or equation 2 to get the distance at which they meet

from equation 1

from equation 2

conduction, convection, radiation

not too sure thinking all accelerations are pointed down