(D) Initial speed of the three-wheeler,$u = 36\; km/h = 36 \times \frac{5}{18} = 10\; m/s$.Final speed of the three-wheeler,$v = 0\; m/s$.Time taken t

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(D) Initial speed of the three-wheeler,$u = 36\; km/h = 36 \times \frac{5}{18} = 10\; m/s$.Final speed of the three-wheeler,$v = 0\; m/s$.Time taken to stop,$t = 4.0\; s$.Total mass of the system,$M = 400\; kg + 65\; kg = 465\; kg$.Using the first equation of motion,$v = u + at$,the acceleration $a$ is:$a = \frac{v - u}{t} = \frac{0 - 10}{4} = -2.5\; m/s^2$.The negative sign indicates retardation.Using Newton's second law of motion,the retarding force $F$ is:$F = M \times |a| = 465 \times 2.5 = 1162.5\; N$.

Class 11 Physics Laws of Motion Second Law of Motion

Medium

The driver of a three-wheeler moving with a speed of $36\; km/h$ sees a child standing in the middle of the road and brings his vehicle to rest in $4.0\; s$ just in time to save the child. What is the average retarding force (in $N$) on the vehicle? The mass of the three-wheeler is $400\; kg$ and the mass of the driver is $65\; kg$.

A
$465$
B
$2312.5$
C
$823.36$
D
$1162.5$

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Laws of Motion

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Second Law of Motion

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