Question

A pickup 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?

Answer 1

Answer:

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. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

$v_{i}$ = 33.2 m/s, a = 0 (since the velocity is constant), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

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

Hatchback:

$a=5m/s^{2}$, $v_{i}$ = 0 m/s (since initial velocity is zero), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

putting in the data we will get

$s=(1/2)(5)t^{2}$

now putting 's' value from eq (1)

$2.5t^{2}-33.2t=0$

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.

Answer 2

Answer:

• 440.9 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 + $0.5at^{2}$

distance covered by the pickup = ut + $0.5at^{2}$

where

• u = initial speed of the pickup = 33.2 m/s
• t = time it takes
• a= acceleration of the pickup = 0
• the distance covered by the pickup (s) now becomes = 33.2t +$0.5.(0).t^{2}$ = 33.2t ...equation 1

       

distance covered by the hatchback = ut + $0.5at^{2}$

where

• u = initial speed of the hatchback = 0 m/s
• t = time it takes
• a= acceleration of the hatchback = 5 m/s^{2}
• the distance covered by the hatchback (s) now becomes = (0)t + $0.5x5t^{2}$

        =  $2.5t^{2}$......equation 2

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

• distance covered by the pickup = distance covered by the hatchback
• equation 1 = equation 2

• 33.2t = $2.5t^{2}$
• 33.2 = 2.5t
• t = 13.28 s

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

• distance (s) = 33.2t = 33.2 x 13.28= 440.9 m

from equation 2

• distance (s) =  $2.5t^{2}$ = $2.5x12.8^{2}$ = 440.9 m