42.5 m far the car would skid with locked brakes at speed of 122.5 km/h.
Explanation:
As per the given question, friction force is constant
A car moving at 49 km/h skids 17 m
[tex]\text { So speed } S_{1}=49 \mathrm{km} / \mathrm{h}[/tex] and
[tex]\text { Distance } \mathrm{D}_{1}=17 \mathrm{m}[/tex]
The car skid with locked brakes at 122.5 km/h "x" distance.
[tex]\text { So speed } \mathrm{S}_{2}=122.5 \mathrm{km} / \mathrm{h}[/tex]
Distance [tex]\mathrm{D}_{2}=\text { unknown }(\mathrm{x})[/tex]
We know that ratio of distance and speed is
[tex]\frac{d_{1}}{d_{2}}=\frac{s_{1}}{s_{2}}[/tex]
[tex]\frac{17}{d_{2}}=\frac{49}{122.5}[/tex]
Cross multiply
[tex]d_{2}=\frac{17 \times 122.5}{49}[/tex]
[tex]\mathrm{d}_{2}=\frac{2082.5}{49}[/tex]
[tex]\mathrm{d}_{2}=42.5 \mathrm{m}[/tex]
The distance skid by the car with locked brakes at speed of 122.5 km/h is 42.5 m.
