Question

4.On applying brakes a vehicles takes 10 seconds to stop ,if it was running at a speed of 40m/s then how long distance will covered it before it stop

Answer 1

Given :

▪ Stopping time = 10s

▪ Initial velocity = 40mps

To Find :

▪ Stopping distance of vehicle.

Solution :

✒ First we have to find out retardation of vehicle after that we can calculate stopping distance with the help of kinematics formula.

✴ First equation of kinematics :

☞ v = u + at

✴ Third equation of kinematics :

☞ v^2 - u^2 = 2as

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✳ Retardation of vehicle :

▪ v = u + at

▪ 0 = 40 - 10a

▪ a = -40/10

▪ a = -4m/s^2

✒ Negative sign shows retardation.

✴ Stopping distance of vehicle :

▪ v^2 - u^2 = 2as

▪ (0)^2 - (40)^2 = 2(-4)s

▪ 1600 = 8s

▪ s = 1600/8

▪ s = 200m

Answer 2

Aɴꜱᴡᴇʀ

☞ Distance travelled by the body is 200m

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Gɪᴠᴇɴ

➳ Initial velocity (u) = 40 m/s

➳ Final velocity (v) = 0 m/s

➳ Time (t) = 10 seconds

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Tᴏ ꜰɪɴᴅ

➜ Distance (s) covered by the body before comming to a stop?

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Sᴛᴇᴘꜱ

❍ The answer to this question can simply be found with the help of the 3rd Law of motion, but for that we first need the value of the acceleration (here retardation) of the body, so it can be found with the help of the First Law of motion,that is,

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

So substituting the given values,

$\leadsto \sf{}0 = 40 + a \times 10 \\ \\ \leadsto \sf{} - 40 = 10a \\ \\ \leadsto \sf \cancel\frac{ - 40}{10} = a \\ \\ \leadsto \sf{ \pink{a = - 4 \: m /{s}^{2}}}$

❍ So as we have the value of acceleration let's calculate the distance covered with the help of,

$\large\underline{\boxed{ \sf\red{{v}^{2} = {u}^{2} + 2as }}}$

Substituting the given values,

$\dashrightarrow \sf{0}^{2} = {40}^{2} + 2 \times ( - 4) \times s \\ \\ \dashrightarrow{} \sf( - 1600 ) = ( - 8s) \\ \\ \dashrightarrow{} \sf \frac{ - 1600}{ - 8} = s \\ \\ \pink{ \dashrightarrow \sf{}s = 200 \: m}$

Note:-

☆ It is not compulsory that to find the distance travelled by the body,we should only use the third law of motion, we can also use the second law of motion, that is S = ut + ½ at²

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