Question

A car is travelling with a velocity of 20m/s suddenly car driver sees a child at a dustance of 50m from his ca whay will be the minimum deaccelaration

Answer 1

Hola User!!

Your answer :-

$given = > \\ distance = s = 50m \\initial \: velocity = u = 20 \frac{m}{s} \\ final \: velocity = v = 0 \frac{m}{s} \\ \\ the \: three \: equations \: of \: motion \: are = \\ = > v = u + at \\ = > s = ut + \frac{1}{2} a {t}^{2} \\ = > {v}^{2} = {u}^{2} + 2as \\ \\ this \: can \: be \: solved \: by \: the \: 3rd \: equation. \\ \\ = > {v}^{2} = {u}^{2} + 2as \\ \\ = > {0}^{2} = {20}^{2} + 2 \times a \times 50 \\ \\ = > 0 = 400 + 100a \\ \\ = > - 400 = 100a \\ \\ = > \frac{ - 400}{100} = a \\ \\ = > - 4 \frac{m}{ {s}^{2} } = a$

But we had to find deceleration so =>

a = - 4
- a = 4m/s^2


Hope it helps!!

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