Answer:
If one over takes the other, then time by the both are same, = t
If we consider the length of the cars, the difference of distance travelled by the both cars \[=5+5=10m\]
Velocity of car \[A\,\,\And \,\,{{V}_{A}}=60km/h\]
\[=60\times \frac{5}{18}=\frac{50}{3}m/s\]
Velocity of car B, \[{{V}_{B}}=42km/hr\]
\[=42\times \frac{5}{18}=\frac{35}{3}m/s\]
distance travelled by car A
\[{{S}_{A}}={{V}_{A}}\times t\frac{50}{3}t\]
distance travelled by car B
\[{{S}_{B}}={{V}_{B}}\times t\frac{35}{3}t\]
\[\therefore \frac{50}{3}t-\frac{35}{3}t=10\Rightarrow t\left[ \frac{50}{3}-\frac{35}{3} \right]=10\]
Since the fastest car travelled more distance, the total road distance used for the overtake
\[={{V}_{A}}\times t+5m=\frac{50}{3}\times 2+5=38.33m=38m\]
You need to login to perform this action.
You will be redirected in
3 sec
