question_answer

Find the ratio of the force acting on them from the velocity-time graph. On turning a comer, a motorist rushing at 36 km/hr, finds a child 51 m ahead. He stops the car within one metre of the child by the application of brakes. Calculate the retarding force and the time required to stop the car. The total weight of the car and passenger is 2000 kg.

Answer:

Initial velocity of a car\[=u=36\,km/hr=\] \[36\times \frac{1000}{3600}m/s=10m/s\] Final velocity of a car\[=v=0\,m/s\] The car stops at a distance = s \[=51\,m-1\,m=50\,m\] Total weight of the car and the passenger = 2000 kg Retarding force\[=F=?\] Time required to stop the car\[=t=?\] We know that F = ma How to find a in terms of the given data u, v, a and t ? We know that \[{{v}^{2}}-{{u}^{2}}=2as\]                             ……(1) Substituting the above values in (1), we get \[{{(0)}^{2}}-{{(10)}^{2}}=2\times a\times 50\Rightarrow -100=100\times a\]c \[\Rightarrow a=\frac{-100}{100}=-1m/{{s}^{2}}\]                                        ….… (2) \[\therefore \]Retarding force \[=F=ma\]                        …… (3) Substituting value of ‘m’ and ‘a’ in (3), we get \[F=(2000\times -1)kgm/{{s}^{2}}\Rightarrow F=-2000N\] Also we know that, \[a=\frac{v-a}{t}\Rightarrow t=\frac{v-u}{a}\]                                    …. (4) Again, substituting the above values in (4), we get \[t=\frac{(0-10)}{-1}=\frac{-10}{-1}\sec =10\sec \] Therefore, time required to stop the car is 10 sec.