Question

What extra force should an engine of a car develop , such that it's velocity Changes from 18km-¹ to 72kmh-¹ over a distance of 20 m when the mass of car is 900 kg?
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Answer 1

Given:

$•u (initial \: vilocity)=18km {h}^{ - 1} \\ = 18 \times \frac{5}{18} m {s}^{ - 1} \\ = 5m {s}^{ - 1}$

$•V (final \: vilocity)=72km {h}^{ - 1} \\ = 72 \times \frac{5}{18} \\ = 4 \times 5 \\ = 20 {ms}^{ - 1}$

$•S(distance) = 20m$

$•m (mass) = 900kg$

To find accerlation:

Formula used:

$</strong><strong>\</strong><strong>b</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>{</strong><strong> {V}^{2} - {u}^{2} = 2as</strong><strong>}</strong><strong>$

Answer:

→$(20)²-(5)²=2×a×20$

→$</strong><strong>400 - 25 = 40a$

→$</strong><strong>375 = 40a$

→$</strong><strong>a = \</strong><strong>d</strong><strong>frac{375}{40}$

→$</strong><strong>\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong> </strong><strong>{</strong><strong>a</strong><strong>= 9.375</strong><strong>✓</strong><strong>}</strong><strong>$

To find extra force:

Formula used :

$</strong><strong>\</strong><strong>b</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>{</strong><strong>Extra\</strong><strong>:</strong><strong>force</strong><strong> = mass \times </strong><strong>ac</strong><strong>c</strong><strong>erlation}</strong><strong>$

Answer:

→$</strong><strong>force=900×9.375$

→$</strong><strong>\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>force</strong><strong> = 8437.5</strong><strong>✓</strong><strong>}</strong><strong>$

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