Question

With what acceleration must the breaks be applied to stop a vehicle travelling with a speed of 40km/he in 20 seconds

Answer 1

Answer:

-0.555 m/s^2

Explanation:

According to the given question,

u = initial velocity = 40 km/h = 40×1000/3600 = 11.1 m/s

t = time = 20 sec

v = final velocity = 0 m/s

a = acceleration and -a = deceleration

And we are asked to find the retardation or deceleration of the vehicle.

We know,

• v = u+at

Putting the values we get

• 0 = 11.1+ 20 a

• 20a = -11.1

• a = -11.1/20

• a = - 0.555 m/s^2

Thus declaration of the vehicle should be 0.5 m/s^2 (approx) in 20 sec to stop the vehicle traveling with 40 km/h .

Means acceleration should be -0.5 m/s^2.

Answer 2

Answer:

Hello Dear User__________

Here is Your Answer...!!

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Step by step solution:

$Given \ u=40km \ hr^{-1} \ and \ t=6 \ sec\\\\we \ have \ to \ find \ a( \ acceleration \ )\\\\First \ let \ us \ convert \ km/hr \ to \ m/sec\\\\40 \ km/hr=\frac{40 \times1000}{60 \times60}=11.11 \ m/sec\\\\From \ first \ equation \ of \ motion \ we \ have\\\\v=u+at\\\\put \ the \ value \ here\\\\0-u=at\\\\-11.11=20a\\\\a=\frac{-11.11}{20}=-0.55 \ m \ sec^{-2}\\\\So \ a=-0.5 \ m \ sec^{-2}\\\\Here \ -a \ means \ declaration\\\\Hence \ -0.5 \ m \ sec^{-2} \ must \ applied \ to \ stop \ the \ vehicle$

Hope it is clear to you.