Question

an object of man 5kg moving in straightline with velocity 5m/s after 2minute it stoped what is the distance travel by the object

Answer 1

Heya!

In the question, it is given that

• Mass of man(m) = 5kg
• Initial velocity(u) = 5 ms^-1
• Final velocity(v) = 0 ms^-1

.....because stopped.....

• Time(t) = 2 min = 2*60 = 120 sec.

.......now to find distance(s), we need to calculate accelaration(a).......

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

$a = \frac{0-5}120\implies a = \frac{-5}120 = -0.416 \: ms^{-2}$

Now, by second equation of motion,

$s = ut + \frac{1}2 at^2\implies$

s = 5(120) + 1/2(-0.0416)(120)(120)

tejasgupta

s = 600 + (-5.77) = 600 - 5.77 = 594.23 meters approx......