question_answer

The driver of a car travelling at 52 km/h applies the brakes and decelerates uniformly. The car stops in 5 seconds. Another driver going at 34 km/h applies his brakes slower and stops after 10 seconds, hi the same graph, plot the speed versus time graph of the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer:

The data given in this numerical problem are in different units. So, we should first convert km/h unit into m/s unit. For first car: Initial velocity\[u=52km/h\] \[=52\times \frac{5}{18}=14.4m/s\] Final velocity, \[v=0km/h=.0.0m/s\] Time taken, \[t=5s\] For second car: Initial velocity \[u=34km/h=34\times \frac{5}{18}=9.4m/s\] Final velocity, \[v=0km/h=.0.0m/s\] Time taken, \[t=10s\] Graph plotting: Draw x-axis and y-axis on a graph sheet. Mark the origin as (0, 0) and plot the speed time graphs of the two cars as shown in figure. Then, the straight lines AB and CD describe the speed-time graphs of the first and the second car respectively. The distance travelled by a moving body is given by the area under its speed-time graph. So, Distance travelled by the first car = Area of the triangle AOB \[=\frac{1}{2}\times OB\times AO\] (\[\therefore \]Area of a triangle\[=\frac{1}{2}\times Base\times Height)\] \[=\frac{1}{2}\times 14.4m/s\times 5s=\frac{1}{2}\times 14.4\times 5m=36m\] Similarly, Distance travelled by the second car = Area of triangle COD \[=\frac{1}{2}\times OD\times CO=\frac{1}{2}\times 9.4m/s\times 10s\] \[=\frac{1}{2}\times 9.4\times 10m=47m\] Thus, the second car travels 47 m and the first car travels 36 m before coming to rest. So, the second car travelled farther after the brakes were applied. Note: When the speed of a body increases or decreases uniformly and its initial or the final speed is not zero, then the area of the trapezium formed by the speed-time graph for the given time interval is equal to the distance travelled by the body in that time interval.