Two trucks leave a warehouse at the same​ time, traveling in opposite directions. The rate of the faster truck exceeds that of the slower truck by miles per hour. After ​hours, they are miles apart. What are the rates of the​ trucks? Use the table below to organize your work

Speed of the slower truck = 57.5 miles/hour

Speed of the faster truck = 62.5 miles/hour

Step-by-step explanation:

Using missing information;

Assume;

Speed of the slower truck = a

Speed of the faster truck = a + 5

Distance traveled by the slower truck = d

Distance traveled by the faster truck = 600 - d

So,

a(5) = d Eq1

(a+5)(5) = 600 - d

5a + 25 = 600 - d

So,

5a + 25 = 600 - 5a

10a = 575

a = 57.5

Speed of the slower truck = 57.5 miles/hour

Speed of the faster truck = a + 5

Speed of the faster truck = 57.5 + 5

Speed of the faster truck = 62.5 miles/hour

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

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