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

• 440.9 m

Step-by-step 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

Answer 2

The pickup truck and hatchback will meet again at 440.896 m

Step-by-step 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.