Respuesta :
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
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
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.
