Question

A boy playing on a ladder and slide starts from rest from the top of the ladder and reaches down in 3 second if the find velocity was 2 metre per second calculate the uniform acceleration produced

Answer 1

Answer:

✡$a=\frac{2}{3}m/s²\:\:\: Or\:\: 0.66m/s²$

Explanation:

Given⤵

• u=0m/s (he slide start from rest)
• t= 3sec
• v= 2m/s

To find ⤵

• a= ?

Solution⤵

Using 1st equation of motion

v=u+at we have,

$a=\frac{v-u}{t}$

Putting the value of u,v and t.

$a=\frac{2-0}{3}$

$a=\frac{2}{3}$

$a=\frac{2}{3}m/s²\:\:\:Or\:0.66m/s²$

Extra points⤵

✡1st equation of motion

v=u+at

✡2nd equation of motion

$s=ut+\frac{1}{2}at²$

✡3rd equation of motion

v²=u²+2as

Where u=initial velocity, v=final velocity, a= acceleration , t= time and s= height or distance

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Hope this is helpful to you!