Question

Acar has initial velocity 72km it acceleration at2m calculate final velocity and distance cover by 3min​

Answer 1

v=u+at

v=72+2×3

v=72+6

v=78km

Answer 2

Aɴsᴡᴇʀ

we have,

initial velocity, u = 72 km/h

first we have to convert the unit of velocity into m/s

$\bold{\implies u = \frac{ 72 \times 1 \cancel{000}}{36 \cancel{0 0}} m {s}^{ - 1} } \\ \\ \bold{\implies u = \frac{ \cancel{72} \times 10}{ \cancel{36}} \:m {s}^{ - 1} \: \: \: \: } \\ \\ \bold{ \implies u = 2 \times 10 \: m {s}^{ - 1} \: \: \: \: \: \: \:} \\ \\ \bold{\implies u = 20 \: m {s}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

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so, now we have,

initial velocity, u = 20m/s

acceleration, a = 2 m/s²

time taken, t = 3 sec

final velocity, v = ?

distance, s = ?

now, by first equation of motion :

$\boxed{ \red {\bold{at = v - u}}}$

$\bold{ : \implies 2 \times 3=v - 20 } \\ \\ \bold{ : \implies6 = v - 20 } \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: v = 6 + 20} \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: \boxed{ \pink {\bold{v = 26 \: m {s}^{ - 1} }}}}$

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now, by second equation of motion:

$\boxed{ \red{\bold{s = ut + \frac{1}{2} a {t}^{2} }}}$

$\bold{ : \implies s =20 \times 3 + \frac{1}{ \cancel{2}} \times \cancel{ 2} \times {3}^{2} } \\ \\ \bold{ : \implies s = 60 + 9} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: \boxed{ \pink{\bold{ s = 69 \: m}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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hence, final velocity of the car is 26 m/s and distance covered by the car is 69 m

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☯︎ important :-

first equation of motion : at = v - u

second equation of motion : s = ut + ½ at²

third equation of motion : v² = u² + 2as