Question

2) The speed of a body moving along the straight path
decreases from 10mls to 5mls in 2s. Find the
decaleration.​

Answer 1

Solution :-

Since the body moving in a straight path . So , the velocity of body is equal to speed . So ,

• Initial velocity = Initial speed = 10m/s
• Final velocity = Final speed = 5 m/s

Now finding deceleration using first equation of motion.

♦ v = u + at

Where ,

• v is final Velocity
• u is Initial velocity
• a is acceleration
• t is time

Substituting the known values we have ,

⇒ 5 = 10 + a × ( 2 )

⇒ 5 - 10 = 2a

⇒ - 5 = 2a

⇒ a = -5/2

⇒ Acceleration = - 2.5m/s²

Here acceleration is negative , so deceleration will be positive . So , deceleration = 2.5 m/s²

Hence , deceleration of the body = 2.5 m/s²

Answer 2

$\underline{\bigstar\:\textsf{Given:}}$

• Initial speed or initial velocity (u) = 10 m/s
• Final speed or final velocity (v) = 5 m/s
• Time (t) = 2 sec

$\underline{\bigstar\:\textsf{To\:find:}}$

• Deceleration or retardation of that moving body (-a) = ?

$\underline{\bigstar\:\textsf{Solution:}}$

$\text{\small\underline{\green{Let's\:take\:analysis\:!}}}$

Here, a body is moving along a straight path and its velocity or speed is decreasing from 10 m/s to 5 m/s in 2 seconds. We have to calculate the magnitude of deceleration(-ve acceleration).

$\text{\small\underline{\green{Let's\:find\:now\:!}}}$

We remember our 1st equation of motion also known as newton's first kinematic equation i.e, v = u + at

(just put the given values to know the amount of deceleration)

$\implies$ 5 = 10 + a × 2

$\implies$ 5 - 10 = 2 × a

$\implies$ -5 = 2a

$\implies\dfrac{-5}{2}$ = a

$\implies$ -2.5 m/s² = a

${\large{\boxed{\sf{\blue{\therefore \: Deceleration \: = 2.5\:m/s^2}}}}}$

Hope it helped you dear...