Physics, asked by rakinanwar2862, 10 months ago

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12 m/s. Its velocity decreases at a uniform rate of 0.5 m/s2 . a) Find the time it will take to come to rest. b) Find the distance covered by it before coming to rest.

Answers

Answered by Anonymous
71

Given :

▪ Initial velocity = 12m/s

▪ Deceleration = 0.5m/s²

To Find :

▪ Time taken by particle to come to rest.

▪ Distance travelled by particle before it is brought to rest.

Concept :

☞ Since, acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

First equation of kinematics :

\bigstar\:\underline{\boxed{\bf{\red{v=u+at}}}}

Third equation of kinematics :

\bigstar\:\underline{\boxed{\bf{\blue{v^2-u^2=2as}}}}

Calculation :

Calculation of time :

\dashrightarrow\sf\:v=u+at\\ \\ \dashrightarrow\sf\:0=12-(0.5)t\\ \\ \dashrightarrow\sf\:t=\dfrac{12}{0.5}\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{t=24\:s}}}}\:\gray{\bigstar}

Calculation of distance :

\Rightarrow\sf\:v^2-u^2=2as\\ \\ \Rightarrow\sf\:(0)^2-(12)^2=2(-0.5)s\\ \\ \Rightarrow\sf\:-144=(-1)s\\ \\ \Rightarrow\underline{\boxed{\bf{\purple{s=144\:m}}}}\:\gray{\bigstar}

[Note : -ve sign of a shows retardation.]


BrainIyMSDhoni: Great :)
Answered by MystícPhoeníx
128

Given:-

  • Initial velocity (u) = 12 m/s .

  • Final velocity (v) = 0 m/s

  • Acceleration (a) = - 0.5 m/s²

To Find:-

  • (a) Time taken to come to rest.

  • (b) Distance covered by before coming to rest.

Solution :-

In First case we have to find the time taken to come to rest.

By using 1st equation of motion

☛ v = u + at

putting all the values which is given above

➭ 0 = 12 + (- 0.5 )× t

➭ -12 = (-0.5 (× t

➭ t = 12/0.5

➭ t = 24 s.

∴ The time taken to come to rest is 24 s .

Now, In Second case we have to calculate the distance covered by it before coming to rest.

Using the 2nd equation of motion

☛ s = ut +1/2at²

➭ s = 12 ×24 + 1/2 × (-0.5 )× 24²

➭ s = 288 +1/2 ×(-0.5 )× 576

➭ s = 288 + (-0.5) × 288

➭ s = 288+(-144)

➭ s = 288 - 144

➭ s = 144 m

∴ The distance covered by it before coming to rest is 144m .

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