Question

A man loses 50% of his velocity after running a distance of 100 m. If his retardation is uniform, the distance he will cover before coming to rest is

• A. $45.2$ m
• B. $33.3$ m
• C. $27.5$ m
• D. $15.7$ m
• E. $50.5$ m

Hint:

When dealing with uniformly accelerated motion, use kinematic equations to find unknowns like distance or acceleration.

Answer 1

The Correct Option is B

To solve the problem, we analyze the motion of the man under uniform retardation. 
Step 1: Define the variables 
- Let the initial velocity of the man be \( u \). 
- After running a distance of \( 100 \, {m} \), his velocity becomes \( \frac{u}{2} \) (since he loses 50% of his velocity). - Let the uniform retardation be \( a \). 
Step 2: Use the equation of motion The equation of motion relating velocity, distance, and acceleration is: \[ v^2 = u^2 + 2as \] where:
\( v \) is the final velocity,
\( u \) is the initial velocity,
\( a \) is the acceleration (or retardation, if negative),
\( s \) is the distance. 
For the first part of the motion (when the man loses 50% of his velocity):
Initial velocity \( = u \),
Final velocity \( = \frac{u}{2} \),
Distance \( = 100 \, {m} \).
Substitute these values into the equation: \[ \left(\frac{u}{2}\right)^2 = u^2 + 2a(100) \] \[ \frac{u^2}{4} = u^2 + 200a \] Rearrange to solve for \( a \): \[ \frac{u^2}{4} - u^2 = 200a \] \[ -\frac{3u^2}{4} = 200a \] \[ a = -\frac{3u^2}{800} \] 
Step 3: Find the total distance before coming to rest Now, we need to find the total distance \( S \) the man covers before coming to rest. 
At rest, the final velocity \( v = 0 \). 
Using the same equation of motion: \[ v^2 = u^2 + 2aS \] Substitute \( v = 0 \), \( u = u \), and \( a = -\frac{3u^2}{800} \): \[ 0 = u^2 + 2\left(-\frac{3u^2}{800}\right)S \] \[ 0 = u^2 - \frac{3u^2}{400}S \] Solve for \( S \): \[ \frac{3u^2}{400}S = u^2 \] \[ S = \frac{u^2 \cdot 400}{3u^2} \] \[ S = \frac{400}{3} \] \[ S \approx 133.33 \, {m} \] 
Step 4: Subtract the initial 100 m The man has already covered \( 100 \, {m} \) before his velocity reduces to \( \frac{u}{2} \). 
The additional distance he covers before coming to rest is: \[ S_{{additional}} = S - 100 \] \[ S_{{additional}} = \frac{400}{3} - 100 \] \[ S_{{additional}} = \frac{400}{3} - \frac{300}{3} \] \[ S_{{additional}} = \frac{100}{3} \] \[ S_{{additional}} \approx 33.33 \, {m} \] 
Final Answer: The distance the man will cover before coming to rest is: \[ \boxed{33.33 \, {m}} \]