A Person is running at a …
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Naveen Sharma 7 years ago
Ans. Let Train cover a distance d before man caught it in t time.
So, Distance covered by man S = 6+d m
Initial Speed of man u = 4m/s
As man is running with constant speed, So a = 0
Using
{tex}S = ut +{1\over 2}at^2{/tex}
{tex}=> 6+d = 4t +{1\over 2}\times 0\times t^2{/tex}
{tex}=> 6+d = 4t +0{/tex}
{tex}=> d = 4t -6 {/tex}………(1)
Similarly initial Speed of Train u = 0
acceleration a = 1m/s2
Distance traveled = d
So,
{tex}=> d = 0\times t+{1\over 2}\times 1\times t^2{/tex}
{tex}=> d ={ 1\over 2} t^2{/tex}………(2)
From (1) & (2)
{tex}=> 4t-6 = {1\over 2}t^2{/tex}
{tex}=> 8t-12 = t^2{/tex}
{tex}=>t^2- 8t+12 =0{/tex}
On solving we get,
t = 6s or 2s
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