Question

A body moves with a velocity 30m/s applies breaks and stop in 20 sec. Find deacceleratiom and displacement covered (for class 9th)​

Answer 1

Answer:

1.5 m/s^2 and 300 m

Explanation:

According to the given question,

u = initial velocity = 30 m/s

v = final velocity = 0 m/s

t = time = 20 sec

a = acceleration and -a = deceleration

s = displacement

And we are asked to find the displacement and retardation.

We know,

• v = u+at

Putting the values we get,

• 0 = 30 + 20a

• 20a = - 30

• a = - 30/20

• a = - 3/2

• a = - 1.5 m/s^2

Thus acceleration of the body is -1.5 m/s^2 or we can say retardation is 1.5 m/s^2.

Again,

• s = ut + 1/2 at^2

Putting the values we get,

• s = 30 × 20 - 1/2 × 3/2 ×400

• s = 600 - 1200/4

• s = 600 - 300

• s = 300 m

Thus displacement of the body is 300 m and deceleration or retardation is 1.5 m/s^2.

Answer 2

Answer:

$\fbox {a = - 1.5 m/s^2}$

$\fbox {Distance (s) = 300m}$

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Given data :-

Initial velocity (u) = 30m/s

Final velocity (v) = 0

Since the body comes to rest after applying breaks , so final velocity is " zero "

Time taken :- 20 s

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Question :- To find deceleration and distance covered

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FOR Solving Problem Like THESE We Use 3 Equations Of Uniform Acceleration .They Are

(1) V = u + at

(2) v^2 - u^2 = 2as

(3) S = ut + 1/2 at^2

We use equation (1) to find deceleration because the rest 2 equations has " s " in it ( s is unknown value )

$\fbox {V = u + at}$

0 = 30 + a (20)

30 + 20a =0

20 a = -30

a = -30/20

a = - 3/2

a = -1.5 m/s^2

Now it's time to calculate " distance"

$\fbox {V^2 - u^2 = 2as}$

0 - (30)^2 = 2 (-1.5)s

- 900 = -3S

900 = 3s

S = 900/3

S = 300m

◇ THING TO KNOW :-

Acceleration in negative direction is called as deceleration.