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 = \(120 km/h + a \times 4.2 hr\)
a = 28.571 \(km/h^{2}\)
Also, it is known that \(v^{2} = u^{2} + 2as\)
s = \(\frac{v^{2} - u^{2}}{2a}\)
Putting the values into the above formula calculate for 's' as follows.
s = \(\frac{v^{2} - u^{2}}{2a}\)
= \(\frac{(0)^{2} - (120)^{2}}{2 \times 28.571 km/hr^{2}}\)
= 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.
\(\frac{252 km}{45 L}\)
= 5.6 km/L
For car Q, v = u + at
0 = \(120 + a \times 5.4 hr\)
a = \(\frac{120 km/hr}{5.4 hr}\)
= 22.22 \(km/hr^{2}\)
Now, calculate value of 's' for car Q as follows.
s = \(\frac{v^{2} - u^{2}}{2a}\)
= \(\frac{(0)^{2} - (120)^{2}}{2 \times 22.22 km/hr^{2}}\)
= 324.0 km
Hence, average kilometers traveled for liter of gas is as follows.
\(\frac{324 km}{45 L}\)
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.
