Physics_Motion_Bacic Sheet, Cheat Sheet of Physics

Physics – Motion Chapter | Basic Sheet (SSC) The Ultimate Basic Sheet for Quick Exam Revision! This Physics – Motion Basic Sheet is designed in a simple, concise, and exam-oriented format to help SSC students understand the chapter easily and revise efficiently. 📚 What's Included? ✅ All important definitions from the Motion chapter ✅ Clear explanations of Distance, Displacement, Speed, Velocity, and Acceleration ✅ Important formulas and SI units ✅ Easy methods for applying formulas ✅ Frequently asked exam concepts and key points ✅ Simple examples and short notes ✅ A smart, revision-friendly layout for quick learning 👨‍🎓 Who Is It For? SSC Science students Students looking for last-minute revision 💡 This Basic Sheet helps you master the core concepts of the Motion chapter in a short time, making your exam preparation faster and more effective. 📥 File Format: PDF 📄 Print-Friendly (A4 Size) Get your copy today and start preparing smarter for your exams!

Typology: Cheat Sheet

2025/2026

Available from 07/01/2026

jaman-ahmed
jaman-ahmed 🇧🇩

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1. Rest
2. Motion
3. Types of motion
4. (Scalars) :
5. (Vectors) :
6. (Distance) :
mL]
7. (Displacement) :
[L] (m)
8. (Speed) :
v d
td, t
(ms1) [v] =
[LT1]
9. (Mean Speed) :
10. (Velocity) :
[LT1 ]ms1
11. :
pf3
pf4
pf5
pf8

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1. Rest

2. Motion

3. Types of motion

4. (Scalars) :

5. (Vectors) :

6. (Distance) :

m L]

7. (Displacement) :

[L] (m)

8. (Speed) :

 v

d t d, t

(ms^1 ) [v] = [LT 1 ]

9. (Mean Speed) :

10. (Velocity) :

[LT^1 ] ms^1

332ms^1

12. (Acceleration) : a

13. (Deceleration) :

i. i. ii. [LT^1 ] ii. [LT^2 ] iii. ms  1 iii. ms

 2

i. i.

ii. ii. iii. iii.

iv. iv.

16. (Gravity) :

17. (Gravitational constant) : (1kg) (1m)

G

Nm^2 kg^2 [L

3 M^1 T

 2 ]

18. (Acceleration Due to Gravity) :

‘g’ [LT

 2 ] ms-2^ 9.81ms^2

19. (Laws of Falling bodies) :

(t) (v) v  t (h) (t) h  t^2

(Equation of Motion of Falling bodies) u t v

h v = u + gt, h = ut +

2 gt

v^2 = u^2 + 2gh.

X

(t) Y (s)

5 ms  2

5 ms  2

5 ms  1

50 kmh ^1 50 kmh  1

50

50 kmh^1

t s

v =

s t u t v t = v  u

v  u t

 a =

v  u t

L, T a

 [a] =

[L]

[T

2 ]

 [a] = [LT

 2 ]

 a =

v  u t

u = v = a = t = s = h =

g =

 v = u + at

 s = ut +

at^2

 v^2 = u^2 + 2as

 v =

s t

 s = ^ 

u + v 2 t

 v = u + gt

 h =

u + v

t

 h = ut +

2 gt

2

 v^2 = u^2 + 2gh

5 ms ^1 10 s 45 ms ^1 4 m s^2 20 m s ^1 4 s 4 m s ^1  4 m s^2 2 m s ^2 20 m s ^1 10 s 54 km h ^1 5 s 4 m s ^2 35 m s^1 ; 125 m] 10 m s ^2 80 m 40 m s^1 50 m g = 98 m s ^2 31 3 m s^1 36 kmh ^1 50 s-

 0 20 ms^2 250 m 10 ms ^2 2 km 20 s

2ms ^2 9 m 10 ms ^1 8 ms^1 72 kmh ^1 20 s- 800 m 20 ms ^1 6 147 m 1 5 ms^2 29 ms^1 72 kmh ^1 42  5 ms^2 36 kmh ^1 8 s 25 ms ^2 30 ms^1 160 m

4 ms ^1 7 ms ^1 0  5 ms ^2 0  4 ms ^2

60 s g- 1 s 1s 25 ms ^1 (g = 98 ms ^2 ). 31 89 m 2 55 s

s =

v = u + at

 a =

v  u t =^

0  10 ms^1 50s =^ ^0 20 ms

 2

s = ut +

2 at

2

= 10 ms^1  50 s +

2 (^0 2 ms

 (^2) )  (50 s) 2

= (500  250) m = 250 m  0 20 ms^2 250 m

10 ms ^2 2 km

u = 0 s = 2 km = 2000 m a = 10 ms^2 t =?

s = ut +

2 at

2

2000 m = 0  t +

 10 ms^2  t^2

5 t^2 ms^2 = 2000 m

t^2 =

2000 m 5 ms^2 = 400 s^2

t = 400 s^2 = 20s

20 s

2ms ^2 9 m

10 ms ^1

v = 10 ms^1 s = 9 m a = 2 ms^2 u =?

v 2 = u^2 + 2as 102 m^2 s^2 = u^2 + 2  2 ms^2  9 m u^2 = (100  36)m^2 s^2 u = 64m^2 s^2

 u = 8 ms^1

8 ms^1

72 kmh ^1 20 s-

u = 72 kmh^1 =

ms^1 = 20 ms^1

v = 0 t = 20 s s =?

v = u + at

 a =

0  20 ms^1 20 s =^ ^ 1 ms

 2

s = ut +

2 at

(^2) = 20 ms (^1)  50 s  1 2 ^ 1 ms

 (^2)  (20 s) 2

= 1000 m  200 m = 800 m 800 m 20 ms ^1 6 147 m

u = 20 ms^1 t = 6 s s = 147 m a =? v =?

s = ut +

2 at

2

1 2 at

(^2) = s  ut

 a = 2(s  ut) t^2

2(147 m  20 ms^1  6s) (6 s)^2

=

2(147 120) m 36 s^2 = 15 ms

 2

v = u + at v = 20 ms^1 + 15 ms^2  6 s = 29 ms^1 1 5 ms^2 29 ms^1 72 kmh ^1

u = 72 kmh^1 =

ms^1 = 20 ms^1

s = (42  2)m = 40 m v = 0

a =?

v

2

= u

2

+ 2as

0 = (20 ms^1 )^2 + 2  a  40 m

a = 

(20 ms^1 )^2 2  40 m

400 m^2 s^2 2  40 m  a =  5 ms^2  5 ms^2 36 kmh ^1 8 s 25 ms ^2

u = 36 kmh^1

=

36  1000 m 3600 s = 10 ms

 1

a = 25 ms^2 t = 8 s v =? S =?

v = u + at = 10 ms^1 + 25 ms^2  8s = 30 ms^1 S = ut +

2 at

2

= 10 ms^1  8 s +

2 ^2 5 ms

 (^2)  (8 s) 2

= 80 m + 2 5  32 m = 160 m 30 ms^1 160 m

4 ms ^1 7 ms ^1

0  5 ms ^2 0  4 ms ^2

u 1 = 4 ms^1 u 2 = 7 ms^1 a 1 = 05 ms^2 a 2 = 04 ms^2 t =?

s = u 1 t +

2 a^1 t

2

s = u 2 t +

2 a^2 t

2

u 1 t +

a 1 t^2 = u 2 t +

a 2 t^2

t =

2(u 1  u2) (a 2  a 1 ) =

2(4 ms^1  7 ms^1 ) 0 4 ms^2  0 5 ms^2

=  6 ms^1  0 1 ms^2 = 60 s

60 s

g- 1 s

u =

g 2 ms

 1

h = 0

h = ut 

2 gt

2

g 2 t 

g t^2

g 2 t (1^ ^ t) = 0 t  0 1  t = 0  t = 1s 1s 25 ms ^1

(g = 98 ms ^2 )

u = 25 ms^1 g = 98 ms^2 H =? t =?

v^2 = u^2  2gH 0 = (25 ms^1 )^2  2  9 8 ms^2  H

H = (25 ms^1 )^2 2  9 8 ms^2  H = 3189 m v = u  gt 0 = 25 ms^1  9 8 ms^2  t

t =

25 ms^1 9 8 ms^2 ^ t = 255 s 31 89 m 2 55 s