car Q and z are put though an endurance test to see if they can travel at 120 km/h for 5.0 hours. Each car has a 45-L fuel tank. Car z must stop to refuel after traveling for 4.2 hours. Car Q, however, travels for 5.4 hours before running out of gas. For each car, calculate the average kilometers traveled for each liter of gas (km/L)

As you know

d=v .t
We need to find the distance traveled in 5.4 hout for q car.

d=v×t
=120km/h×5.4h
=648km

so, the total disctance of the car 1 on 35 liters is 648 km
45L=648km

To get the Value for 1 Liter Divide both sides with 45 and you get

45L/45L=648/45L

=14.4km/L

Answer Link

OlaMacgregor OlaMacgregor · Answer 2 · 2019-10-10T09:53:39+02:00

Explanation:

For car Z, according to the Newton equation,

v = u + at

where, v = final velocity = 0

u = initial velocity = 120 km/h

a = acceleration

t = time = 4.2 hours

Putting the given values into the above formula as follows.

v = u + at

0 = [tex]120 km/h + a \times 4.2 hr[/tex]

a = 28.571 [tex]km/h^{2}[/tex]

Also, it is known that [tex]v^{2} = u^{2} + 2as[/tex]

s = [tex]\frac{v^{2} - u^{2}}{2a}[/tex]

Putting the values into the above formula calculate for 's' as follows.

s = [tex]\frac{v^{2} - u^{2}}{2a}[/tex]

= [tex]\frac{(0)^{2} - (120)^{2}}{2 \times 28.571 km/hr^{2}}[/tex]

= 252.0 km

Since, 45 L of fuel is used by the car to travel 252 km. Then calculate car traveled km per liter as follows.

[tex]\frac{252 km}{45 L}[/tex]

= 5.6 km/L

For car Q, v = u + at

0 = [tex]120 + a \times 5.4 hr[/tex]

a = [tex]\frac{120 km/hr}{5.4 hr}[/tex]

= 22.22 [tex]km/hr^{2}[/tex]

Now, calculate value of 's' for car Q as follows.

s = [tex]\frac{v^{2} - u^{2}}{2a}[/tex]

= [tex]\frac{(0)^{2} - (120)^{2}}{2 \times 22.22 km/hr^{2}}[/tex]

= 324.0 km

Hence, average kilometers traveled for liter of gas is as follows.

[tex]\frac{324 km}{45 L}[/tex]

7.2 km/L

Thus, we can conclude that the average kilometers traveled for each liter of gas (km/L) by car Z is 5.6 km/L and for car Q is 7.2 km/L.