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Q. A car moving at $160 \,km/h$ when passes the mark-$A$, driver applies brake and reduces its speed uniformly to $40 \,km/h$ at mark-$C$. The marks are spaced at equal distances along the road as shown below.
image
At which part of the track the car has instantaneous speed of $100 \,km/h$? Neglect the size of the car.

Motion in a Straight Line

Solution:

From $A$ to $C: (40)^2 - (160)^2 = 2a(AC)$
From $A$ to $D: (100)^2 - (160)^2 = 2a(AD)$
Here $D$ is the point where speed is $100 \,km/h$.
From above $AD = \frac{13}{20} (AC)$ and $AB = \frac{AC}{2}$
We see that $AD > AB$.