Equations of Motion (Physics)

Equations of Motion Made Easy! Newton's Equations of Motion also known as SUVAT equations are explained in detail here. We have compared the upward and downward motion.

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[binumuthoor]

ANSWER OF THE FIRST QUESTION EXPLANATION :
1. u {initial velocity} = 0
a {acceleration} = 2 m/s square
t {time} = 5 s
v {final velocity} = ?
So here we have to find the final velocity.
So we take the first equation.
v = u + at
= 0 + 2 x 5
= 10 m/s
So the final velocity is 10 m/s
So the speed or velocity of the car after 5 s is 10 m/s
Please like if helpful...

[jankijoshi]

Q1 - According to first equation of motion
v = u + at
Here,
v = ?
u = 0 m/s
a = -2m/s square
t = 5 s
Put the value in equations
v = 0 m/s + (2 m/s square * 5s)
v = 0 m/s + 10 m/s
Hence speed of the car is 10 m/s.

Q2 - According to third equation of motion
2as = v square - u square
Here,
s = 10 m
v = 0 m/s
u = 72 km/h or 20 m/s
Put the values in the equation
2*a*10 m = 0 square - 20 square
20 m * a = 0 - 400 m/s square
a = -400 m/s square / 20 m
a = -20 m/s square

Q3 - According to the third equation of motion
2as = v square - u square
Here,
a = (-10 m/s square) because the acceleration due to gravity is in upward direction
v = 0 m/s
u = 10 m/s
Put the values in the equation
2 * (-10 m/s square) * s = 0 square - 10 square
-20 m/s square * s = -100 m/s
s = -100 m/s / - 20 m/s square
s = 5 m

2. According to the first equation of motion
v = u + at
Here,
v = 0 m/s
u = 10 m/s
a = -10 m /s square
t = ?
Put the values in the equation
0 = 10 m/s + (-10 m/s square * t )
10 m/s square * t = 10 m/s
t = 10 m/s / 10 m/s square
t = 1s
As the time taken by the ball to cover 5m distance is 1 sec so it will take 1+1 sec = 2 sec to reach the thrower

[benza]

my understanding just went from not having a clue what any of this meant, to being able to get it right most of the time. Great video.

[StaceyMcAllen]

Number 2
Using V2=U2+2as
Convert 72km/h to m/s
=72x1000/60x60= 20m/s

V=0
U=20m/s
S=10
a=?

0^2=(20)^2+2ax10
0=400+20a
20a=-400
a=-400/20
a=-20m/s^2
I hope Im able to help those that were so confused at first like me 🙏

[pixelpower.random]

1. v = u + at = 0 + 2*5 = 10 m/s
2. v2 = u2 + 2as --> 0 = 400 + 20a
a = -400/20 = -20 m/s2
3. t = u/g = 1 s (Single sided journey)
Total time = 2t = 2 s
s = ut + at2/2 = 10 - 5 = 5 m

[shreyashgote-]

Number 2
Using V2=U2+2as
Convert 72km/h to m/s
=72x1000/60x60= 20m/s

V=0
U=20m/s
S=10
a=?

0^2=(20)^2+2ax10
0=400+20a
20a=-400
a=-400/20
a=-20m/s^2
I hope Im able to help those that were so confused at first like me

[abhishekshyleshnair]

Q1) u= 0
a=2
t=5
v=?
(1)v=u+at= 0+(2 x 5)= 10m/s

Q2) a=?, u=20 m/s(72 km/hr- m/s), v=0, s=10
(3) s= v^2- u^2/ 2a
10= 0^2 - 20^2/ 2x (a)
a=-20^2/ 2x 10
a= -400/200
a= -4m/s

Q3) Use equation v=u +at to find time= 1/20
then use equation s=ut+1/2at^2= 1/4 or 0.25 m

[sharmiliagarwal]

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[GeorgePhiri-nklt]

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[geetanjalimisaal]

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[sophielo]

I have a physic test tomorrow about this topic😭😭😭

[aryanmallick]

1. final velocity (v)= 10m/s
2. acceleration (a) = -20m/s^2
3. i) time (t)= 1s
ii) displacement (s)= 5m

[aniruddhadutta]

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[golmaalediting_]

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[mildredkenalemang]

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[sudarshanbalaji]

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[new-knowledge]

The ball when at the top, it is in motion exactly as much as it is in motion, when it is at the bottom. That is because the ball is always in motion within space-time. All you can do is change its direction of travel of the ball within the 4D environment known as space-time.

[prathameshpatil]

Q.1) Given : initial velocity = 0 m/s, acceleration = 2 m/s^2, time = 5 s. To Find : speed of the car. To find the speed of the car we can apply the 1st equation of motion i.e v = u + at, v = 0 + 2*5, v = 0 + 10, v = 10 m/s. Therefore the speed of the car after 5 s is 10 m/s.
Q.2) Given : Initial velocity = 72 km/hr. Now we have to convert 72 km/hr in m/s i.e 72 * 5/18 = 20 m/s. displacement = 10 m. final velocity = 0 m/s. To find : retardation of the brakes. Now to find the retardation of the brakes we can apply the 3rd equation of motion i.e v^2 - u^2 = 2as, (0)^2 - (20)^2 = 2a(10), 0 - 400 = 20a, - 400 = 20a, - 400/20 = a, - 20 m/s^2 = a. Therefore the retardation of the brakes is - 20 m/s^2.
Q.3) Given : Initial velocity = 10 m/s, acceleration due to gravity = - 10 m/s^2, Final velocity = 0 m/s. To find : height & time taken. Now to find the height reached by the ball we can apply the third equation of motion i.e v^2 - u^2 = 2as, (0)^2 - (10)^2 = 2(-10)s, 0 -100 = -20s, -100/-20 = s, 5 m = s. Therefore the height reached by the ball is 5 m. Now to find the time taken to come back to the thrower we can apply 1st equation of motion i.e v = u + at, 0 = 10 + (-10)t, -10 = -10t, -10/-10 = t, 1 sec = t. Now the time taken for the ball to reach the height of 5 m was 1 sec. So, the time taken for the ball to reach to the person who throwed the ball will be 2 seconds.
Sir I have solved the top three questions kindly tell me the answers are right or wrong.

[sarvajagannadhareddy]

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE.
Procedure
Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3
At the end we get the EXACT circumference of the inscribed circle