Question

An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s.What will be the velocity after 7 s from the start.​

Answer 1

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An object starting from rest travels 20 m in first 2 s and 160 m in next 4s.What will be the velocity after 7 s from the start.

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Here ;

$\sf{ u=0, S =20 m ,T=2 sec}$

$\Large { {\boxed {\underline{ { \sf\: Formula \implies \: s \: = ut + \frac{1}{2} at {}^{2} }}}}}$

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$\large\implies \: \tt 20 = 0 + \frac{1}{2} a \times 4$

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$\large\implies \: \tt \: {\red{ a = 10m/ s {}^{2} }}$

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Now,

Velocity at the end of 2 sec ;

$\Large { {\boxed {\underline{ {{ \sf\:Formula \implies \: v = u + at}}}}}}$

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$\large\implies \tt \: v = 10 \times 2$

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$\large\implies \tt \: \red {\:v = 20m/s }$

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Now,

Checking the motion for next 4 sec, let assume acceleration be a1 as we are not sure about the motion in second part ;

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$\sf {u=20 m/s ,\:S=160 m ,\: T=4 sec}$

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$\Large { {\boxed { \tt\ \: s \: = ut + \frac{1}{2} at {}^{2} }}}$

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$\large\tt{ \implies 160 = 20 \times 4 - \frac{1}{2} a1 \times 16}$

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$\large\implies \tt{\red{ a1 = 10m/s {}^{2} }}$

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So,acceleration is constant in both the motion.

Now,

we can easily calculate velocity as ;

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$\sf {u=0,\: T= 7sec , \: a=10m/s^{2}}$

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$\Large{\boxed{\tt{ \:v = u + at}}}$

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$\large\implies\tt \: v = 0 + 10 \times 7$

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$\large\implies \tt \orange{ v = 70m/s^{2}}$

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