Question

In a competition a sprinter starts from rest and achieves a velocity of 30 m/s in a time of 6s.What is his average acceleration?
a.10 m/s 2
b.8 m/s 2
c.7 m/s2
d.5 m/s2​

Answer 1

Answer:

d. 5m/s²

Explanation:

acceleration ( a) = [final velocity (v) - initial velocity (u) ] / time taken (t)

u = 0 since he had started from rest

v = 30ms-1

t = 6s

a =( v-u)/t

= (30-0) /6

= 30/6

= 5ms-2

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Answer 2

$\large\sf\underline{Given\::}$

• Initial velocity ( u ) = 0 m/s ( since sprinter starts from resting position so initially the velocity is 0 )

• Final velocity ( v ) = 30 m/s

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$\large\sf\underline{To\:find\::}$

• Average acceleration

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$\large\sf\underline{Concept\::}$

From the first Kinematical equation we know :

• $\bf\color{blue}{v=u+at}$

where :

• v = $\sf\color{teal}{final\:velocity}$

• u = $\sf\color{teal}{initial\:velocity}$

• a = $\sf\color{teal}{acceleration}$

• t = $\sf\color{teal}{time}$

So we will solve this question using this equation as the values for u , v , and t are already given . Doing so we can get easily the value of a .

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$\large\sf\underline{Solution\::}$

$\sf\:v=u+at$

• Substituting the values of v , u and t

$\sf:\implies\:30=0+a \times 6$

$\sf:\implies\:30=0+6a$

$\sf:\implies\:30=6a$

$\sf\:oR\:6a=30$

• Transposing the value of 6 to RHS it goes to the denominator

$\sf:\implies\:a=\frac{30}{6}$

• Reducing it to the lower terms

$\sf:\implies\:a=\cancel{\frac{30}{6}}$

$\small{\underline{\boxed{\mathrm\red{:\implies\:a\:=\:5\:m/s^{2}}}}}$ ★

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$\large\sf\underline{Other\:Kinematical\:Equations\::}$

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• $\sf\:s=ut+\frac{1}{2}at^{2}$

• $\sf\:v^{2}=u^{2}+2as$

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!! Hope it helps !!