By raising a position of an object, Its potential energy increases.
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.
= 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.
<em>Answer:</em>
<em>When </em><em>a </em><em>body </em><em>is </em><em>moving </em><em>on </em><em>a </em><em>circle </em><em>it </em><em>is </em><em>accelerating </em><em>because </em><em>centripetal </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>it </em><em>towards </em><em>the </em><em>center.</em>
<em>Please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em>
<em>From </em><em>the </em><em>above </em><em>diagram,</em><em>we </em><em>can </em><em>say </em><em>the </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>the </em><em>body </em><em>when </em><em>it </em><em>moves </em><em>in </em><em>a </em><em>circle.</em>
<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Force is equal to mass multiplied by acceleration, therefore
F=ma
m=2569.6 kg
a=4.65m/s^2
therefore F=2569.6*4.65=11948.6 (correct to 1 d.p.)