Question

a truck moving at a speed of 72 kmph applies brakes and stops after every 5 seconds.the distance covered by the during this time period is A.100m B.25m C.50m D.75m

Answer 1

Answer :

Initial speed = 72km/h = 20m/s

Final speed = zero

Time taken by truck to stop = 5s

We have to find Distance covered by truck before it is brought to rest$.$

First we have to find acceleration of truck.

➝ v = u + at

➝ 0 = 20 + a(5)

➝ 5a = -20

➝ a = -4m/s²

◈ Negative sign of a shows retardation of truck$.$

Now let's caluculate distance :)

⇒ v² - u² = 2as

⇒ 0² - 20² = 2(-4)s

⇒ -400 = -8s

⇒ s = -400/-8

⇒ s = 50m (Option C is correct)

God Bless ☃

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is Option C

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$\huge\sf\blue{Given}$

✭ Initial Velocity = 72 km/h

✭ Final Velocity = 0 m/s

✭ Time = 5 sec

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$\huge\sf\gray{To \:Find}$

◈ The distance travelled by the body?

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$\huge\sf\purple{Steps}$

On converting the speed from km/h to m/s

»» $\sf 72 \ km/h = 72 \times \dfrac{5}{18}$

»» $\sf \green{72 \ km/h = 20 \ m/s}$

Here we shall now find out the acceleration (retardation) of the body, and for that we shall use the first equation of motion, that is,

$\underline{\boxed{\sf v = u+at}}$

Substituting the given values,

➝ $\sf v = u+at$

➝ $\sf 0 = 20+a\times 5$

➝ $\sf -20 = 5a$

➝ $\sf \dfrac{-20}{5} = a$

➝ $\sf \red{a = -4 \ m/s^2}$

Next we shall find the distance travelled with the help of the third equation of motion, that is,

$\underline{\boxed{\sf v^2-u^2 = 2as}}$

➳ $\sf 0^2-20^2 = 2(-4)(s)$

➳ $\sf -400 = -8s$

➳ $\sf \dfrac{-400}{-8} = s$

➳ $\sf \orange{s = 50 \ m}$

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