A metal sphere is sliding without friction down a slated incline, so that its acceleration down the slope is given by $a = + 4.00 m / s^{2}$ . Half a second later, it has moved $1.00 m$ down the slope.
What was its initial velocity?

- 3.00 m / s

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- 1.00 m / s

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0.00 m / s

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+ 1.00 m / s

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+ 2.00 m / s

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Solution

The correct option is B $+ 1.00 m / s$
Given : $a = + 4.00$ $m / s^{2}$ $t = 0.5$ s
Distance covered in 0.5 s, $S = 1.00$ m
Let the initial velocity be $u$ .
Using $S = u t + \frac{1}{2} a t^{2}$

$∴$ $1.00 = u \times 0.5 + \frac{1}{2} \times 4.00 \times (0.5)^{2}$ $⟹ u = + 1.00$ $m / s$

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A metal sphere is sliding without friction down a slated incline, so that its acceleration down the slope is given by a=+4.00m/s2. Half a second later, it has moved 1.00m down the slope.What was its initial velocity?

The correct option is B +1.00m/sGiven : a=+4.00 m/s2 t=0.5 s Distance covered in 0.5 s, S=1.00 mLet the initial velocity be u.Using S=ut+12at2∴ 1.00=u×0.5+12×4.00×(0.5)2 ⟹u=+1.00 m/s

A metal sphere is sliding without friction down a slated incline, so that its acceleration down the slope is given by $a = + 4.00 m / s^{2}$ . Half a second later, it has moved $1.00 m$ down the slope.
What was its initial velocity?

The correct option is B $+ 1.00 m / s$
Given : $a = + 4.00$ $m / s^{2}$ $t = 0.5$ s
Distance covered in 0.5 s, $S = 1.00$ m
Let the initial velocity be $u$ .
Using $S = u t + \frac{1}{2} a t^{2}$

$∴$ $1.00 = u \times 0.5 + \frac{1}{2} \times 4.00 \times (0.5)^{2}$ $⟹ u = + 1.00$ $m / s$