Question

Plz find the answer for this

Answer 1

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$<b><u>Answer</u></b>$

$<b>(A) Acceleration is 0.6m/s^2.</b>$

$<b>(B) Distance is 62.4.</b>$

$<b><u>Step-by-step explanation</u></b>$

$<b>(A) Given that,</b>$

Final velocity (v) = 3 m/s

Intial velocity (u) = 0 m/s

Time (t) = 5 secs.

$<b>To find,</b>$

Acceleration (a) = ?

By using 1st equation of motion.

That is, v = u+at

By substituting the given (or) above values,

3m/s = 0m/s + a × 5sec

3 = a × 5

a = $\frac{3}{5}$

a = $0.6 m/s^2$

$<b>Therefore, the acceleration is 0.6m/s^2.</b>$

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$<b>(B) Given that,</b>$

Final velocity (v) = 0 m/s

Intial velocity (u) = 3 m/s

Time (t) = 8secs.

Acceleration = $0.6m/s^2$ [found in (A)]

$<b>To find,</b>$

Distance (s) = ?

By using the 2nd equation of motion.

That is, $s = ut + \frac{1}{2} at^2$

By the substituting the given (or) above values.

s = $3m/s × 8secs + \frac{1}{2} 0.6m/s^2 × 8 × 8$

s = 24 + 0.6 × 8 × 8

s = 24 + 0.6 × 64

s = 24 + 38.4

s = 62.4 m

$<b>Therefore, the distance is 62.4m</b>$

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