Question

A student releases a marble from the top of a ramp and the marble increases speed steadily and travels 190 cm in 4.10 s what's the marbles final speed

Answer 1

Answer:

The value is  $v = 92.68 \ cm/s$

Explanation:

From the question we are told that

     The distance traveled is $d = 190 \ cm = 1.9 \ m$

     The  time taken is  $t = 4.10 \ s$

Generally from kinematic equation we have that

      $s = ut + \frac{1}{2} at^2$

Here u is the initial velocity of the marble and the value is $u = 0 \ m/s$

So

      $1.90 = 0* t + \frac{1}{2} a (4.10 )^2$

=>   $1.90 = 0* t + \frac{1}{2} a (4.10 )^2$

=>    $a = 0.226 \ m/s^2$

Generally from kinematic equations we have that

     $v = u + at$

=>   $v = 0 +0.226 * 4.10$

=>   $v =0.9268 \ m/s$

Converting to  cm/s

=>    $v =0.9268 * 100$

=>    $v = 92.68 \ cm/s$