Question

A person starts his motion in straughtline at a velocity of 15mpers His velocity is changing at a constant rate if he stops after 75 seconds what is his acceleration

Answer 1

Answer:

-0.2 m/s²

Explanation:

Given :

• Initial velocity = u = 15 m/s

• Final velocity = v = 0 m/s

• Time taken = t = 75 seconds

To find :

• Acceleration of the person who has Initial velocity of 15 m/s and final velocity of 0 m/s and time taken is 75 seconds

Acceleration = (v-u) / t

Acceleration = (0-15)/75

Acceleration = - 15/75

Acceleration = - 1/5

Acceleration = - 0.2 m/s²

The acceleration of that person is equal to - 0.2 m/s²

Answer 2

✯✯ QUESTION ✯✯

A person starts his motion in straughtline at a velocity of 15mpers His velocity is changing at a constant rate if he stops after 75 seconds what is his acceleration?

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✰✰ ANSWER ✰✰

$\implies\tt{Initial\:Velocity(u)=15m/s}$

$\implies\tt{Final\:Velocity(v)=0m/s}$

$\implies\tt{Time\:Taken(t)=75\:sec}$

$\implies\tt{Acceleration(a)=?}$

➥Using 1st Equation : -

$\implies\tt{\small{\boxed{\bold{\bold{\red{\sf{v=u+at}}}}}}}$

➥Putting Values : -

$\implies\tt{0=15+a(75)}$

$\implies\tt{0=15+75a}$

$\implies\tt{0-15=75a}$

$\implies\tt{-15=75a}$

$\implies\tt{a=\cancel\dfrac{-15}{75}}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{a}\orange{=}\purple{-0.2}}}$

☛So , The acceleration is 0.2m/s²..