Atrucks relation to a car is -60. the car and truck move in the same direction. the car moves five times as fast for 8 seconds. where is the truck in relation to the car? what operations would you use to solve this and why?

Explanation:

We are not given units for distance or speed. The only units we're given are for time (seconds). Let the car's speed be represented by "v".

When the truck is moving 5 times as fast as the car, its relative speed is ...

  5v -v = 4v

For some time (t) at the higher speed, it will move a distance farther than the car moves. That distance is the product of the relative speed and the corresponding time period: 4vt. Since the speeds are in the same direction, this additional distance is added to the initial relative location of the truck, -60. The new position of the truck relative to the car is ...

  -60 +4vt = -60 +4v(8 seconds)

  -60 + 32v . . . . . new position of the truck in relation to the car

(This only makes sense if the units of v are the same distance units as -60 and time units of seconds.)

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At least, operations of multiplication and addition are needed. Depending on the units of speed and distance, units conversion may also be required.

c. 6tan80

step-by-step explanation:

hello, sorry being so rude after all that fighting i found it kind of funny, so as me saying sorry here is an answer!

since the table showed her that   2km = 2000m.

to determine, how many km in 2000m, she would divide by 1000.

= 2000 / 1000 = 2 km.

similarly, for 8000m = 8000 / 1000 = 8 km

cheers to savings

~

your pal papaguy