Question

a racing car has a uniform acceleration of 6m/s square .what distance will it cover in 10sec after start​

Answer 1

Answer:

60 m/s²

Explanation:

To Find

velocity =displacement upon time

so velocity is 6m/s

Time is 10second

displacement=?

6 =d upon 10

d =10x6=60

Answer 2

$\huge \bf {\underline{\underline{Answer:-}}}$

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$\dag \cal{\underline{Given:-}}$

$\colon\mapsto\tt {Acceleration(a)=6m/s^{2}}$

$\colon \mapsto \tt{Time\: period(t)=10 seconds}$

★By this, we can say that the car is initially at rest.

$\colon \mapsto \tt{Initial\: velocity (u)=0}$

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$\dag \cal {\underline{To\: find:-}}$

★The distance(S),car will cover in 10 seconds after its start .

$\dag \cal {\underline{kinematic\: equation\: to\: be\: used:-}}$

$\large \implies \purple{S=ut+\frac{1}{2}at^{2}}$

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$\dag \cal {\underline{Where,}}$

$\to {S=Distance\:, the\: car\: will\: cover}$

$\to {u=Initial\: velocity}$

$\to{u="Zero"\: in\: this\: case}$

$\to {T=Time\: Period}$

$\to {a=Acceleration}$

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$\dag \cal {\underline{Substituting\: in\: the\: equation,}}$

$\small \colon \mapsto{S=ut+\frac{1}{2}at^{2}}$

$\small \colon \mapsto{S=0\times10sec+\frac{1}{2}\times6m/s^{2}(10sec)^{2}}$

$\small \colon \mapsto{S=0+\frac{1}{\cancel{2}}_{1}×^{3}\cancel{6}\frac{m}{\cancel{s^{2}}}\times100\cancel{s^{2}}}$

$\small \colon \mapsto{S=3m\times100}$

$\large \to \pink {\fbox {S=300m}}$

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$\dag \cal {\underline{Henceupon,}}$

★The Distance car will cover in 10sec after start is 300m

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