Question

Two cars are travelling towards each other on a straight road at velocities $15m{s}^{-1}$ and $16m{s}^{-1}$ respectively. When they are $150$m apart, both the drivers apply the breaks and the cars decelerates at $3m{s}^{-2}$ and $4m{s}^{-2}$ until they stop. How for apart will they be when they have come to a stop.

• A. $86.5$m
• B. $89.5$m
• C. $85.5$m
• D. $81.36$m

Answer 1

The correct option is D $81.36$m

solving with respect to the car $1$ moving with speed $15m/s$.

using concept of relative motion we get $V_{21}=V_{2g}-V_{1g}$

same with acceleration $a_{21}=a_{2g}-a_{1g}$

so velocity car $2$ is $15+16=31m/s,acc=4+3=7m/s^2$and $separation=150m$

distance covered before coming to rest $\frac{{31}^2}{2\times 7}=S=>S=68.64$

so now separation between them is $150-68.64=81.36m$

so none of the option is correct.