Question

An object moving with a speed of 20km/h accelerates with 4km/h^2 and comes in rest. What distance it has covered.​

Answer 1

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An object moving with a speed of 20km/h accelerates with 4km/h^2 and comes in rest. What distance it has covered.

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= 50 km

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Initial velocity = 20 km/hr

Final velocity= 0 km/hr(body comes to rest).

Accleration = -4 km/hr^2(deceleration)

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Distance covered before stopping..?

$\large{ \sf{ \diamond{ \: \: u \: must \: know....}}} \\ \\ \huge{ \green{ \boxed{ {v}^{2} - {u}^{2} = 2as}}} \\ \large{ \red{ \sf{......2nd \: equations \: of \: motion}}}$

$\large{ \sf{ \underline{according \: to \: question..}}} \\ \\ \large{ \sf{ \implies{ we \: have \: \: \: u = 20 \: km \: {h}^{ - 1} }}} \\ \\ \large{ \sf{ \implies{ \: and \: \: a = - 4 \: km \: hr(deceleration) {}^{ - 2} }}} \\ \\ \large{ \sf{ \green{ body \: comes \: to \: rest}}}... \\ \large{ \sf{ \implies{so \: final \: velocity \: v = 0 \: km \: hr {}^{ - 1} }}} \\ \\ \large{ \sf{ \green{putting \: these \: values \: in \: formula...}}} \\ \\ \large{ \sf{ \implies{(0) {}^{2} - (20) {}^{2} = 2 \times( - 4 )\times s}}} \\ \\ \large{ \sf{ \implies{0 - 400 = - 8s}}} \\ \\ \large{ \sf{ \implies{ - 8s = - 400}}} \\ \\ \large{ \sf{ \implies{s = \cancel{ \frac{ - 400}{ - 8} }}}} \\ \\ \large{ \sf{ \implies{ \red{ \boxed{s = 50 \: km}}}}}$

$\large{ \purple{ \bold{ required \: answer \: is \: 50 \: km}}}$

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3 equations of motion...

• v = u + at

• v² = u² + 2as

• s = ut + ½at²