Question

A mouse is 15 m ahead from a cat.The cat runs at 2 m / s ^ 2 uniform acceleration to catch the mouse, at the same time when the mouse is running at a constant speed of 14 m / s, can the cat catch the mouse?​

Answer 1

Answer:

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Answer 2

Let's assume the speed of the cat and mouse will be same after 't' second.

Formula used,

• $\bf{s=ut+\frac{1}{2}at^2}$
• $\bf{s=vt}$

In case of the cat,

• U = 0 m/s
• a = 2 m /s²

$\sf{Here,\:s_1=ut+\frac{1}{2}at^2}$

$\sf{→s_1=0+\frac{1}{2}.2.t^2}$

$\sf{→s_1=t^2}$

In case of the mouse,

• v = 14 m/s

$\sf{Here,\:s_2=vt}$

$\sf{→s_2=14t}$

According to question,

$\tt{s_1=s_2+15}$

$\tt{\leadsto\:t^2=14t+15}$

$\tt{\leadsto\:t^2-14t-15=0}$

$\tt{\leadsto\:t^2-15t+t-15=0}$

$\tt{\leadsto\:t(t-15)+1(t-15)=0}$

$\tt{\leadsto\:(t-15)(t+1)=0}$

$\sf{Either,\:t-15=0}$

$\sf{→t=15\:s}$

$\sf{Or,\:t+1=0}$

$\sf{→t=-1}$

___________________________

The passed distance of the mouse, $s_2=(14×15)+15$

$s_2=225\:m$

The passed distance of the cat,

$s_1=t^2$

$s_1=15^2=225\:m$

Hence, the cat can catch the mouse.