In my last article, The definition and types of Motion were discussed. Now this is an article to better understand Equations of motion with several questions solved.

**Related Terms**

**Displacement:**It is the distance moved in a specified direction. It is a vector quantity.**Distance:**It is the magnitude of separation between two points. It is a scalar quantity.**Speed:**It is the rate of change of distance with time. It is a scalar quantity.**Velocity:**It is the rate of change of displacement with time. It is a vector quantity.

**Acceleration:**It is the increasing rate of change of velocity with time. It is a vector quantity.**Retardation:**It is the decreasing rate of change of velocity with time.

**Equations of motion**

1)

3)

4)

**Derivation**

**Average speed = s/t**

(V+u)/2 =s/t

where, V =final velocity

U = Initial velocity

S = Distance

T = time

Make "s" subject of the formula.

..........(1)

Average acceleration= change in Velocity/time

a = (v-u)/t

at = (v-u)

V = U + at ...........(2)

Substitute (2) in (1)

S =[ (u + at +u)/2]t.

S =[(2u+at)/2]t

........(3)

From (2)

V = u + at

t =(v-u)/a

Substitute the value of t in (1)

S = (v+u)/2 x (v-u)/a

.....(4)

**Question 1**:

A car travels with a uniform velocity of 30m/s for 5s and then comes to a rest in the next 10s with a uniform deceleration.

Find,

a) deceleration

b) total distance travelled.

Solution,

a) using, V = u + at

Deceleration= -a

0 = 30m/s - a(10)

10a = 30

a = 3m/s/s

b) Let, "Stage A" be the period before deceleration.

"Stage B" be the period after deceleration.

At Stage A,

U= 30m/s

t1= 5s

S1= Distance travelled at stage A ?

Using,

From the V-T graph,

Initial Velocity (U) = S1/T1

S1 = ( 30 x 5) m

S1 = 150 m

At Stage B,

U= 30 m/s

V = 0

T2 = 10s

S2 = Distance travelled at stage B ?

Using,

0 = (30)(30) + 2(-3)(s2)

900 = 6(s2)

S2 =150m

Total distance travelled (S) = S1 + S2

S = (150 + 150) m/s

S = 300m.

**Question 2:**

**A bus moves from rest with an initial acceleration of 2m/s^2 for the first 10 s. It then accelerates at a uniform rate of 1m/s^2 for another 15s. It continues at a constant speed for 70s and finally comes to a rest in 20s by a uniform deceleration. From the graph calculate.**

I)Total distance travelled

II) Average speed for the journey

III) Average retardation as it is brought to rest.

IV) Maximum speed attained by the motion.

Solution,

From the V-T graph

I). for the first 10s,

U= 0

a = 2m/s/s

t1 = 10s

Velocity (V)= u + at

V = 0 + (2)(10)

V = 20m/s

Distance (S1) = [(v +u)/2] t

S1 = [ (20 +0)/2] x 10

S1 = [10 x 10] = 100m

For the second 15s

U= 20m/s

t2 = 15s

Velocity (v) = ( 20 + 15) m/s

V = 35 m/s

Distance (S2) = [(v +u)/2] t

S2 = [ (35 +20)/2] x 10

S2 = [41.25 x 10] = 412.5m

For the third 70s,

V = constant at 35m/s

t3 = 70s

V = S3/t3

S3 = (35 x 70) m

S3 = 2450m

For the last 20s (Retardation or deceleration).

V = 0

U = 35m/s

t4 = 20s

S4 = [(v +u)/2] t4

S2 = [ (0 +35)/2] x 20

S2 = [35 x 10] = 350m

Total distance travelled(S)= S1 + S2 + S3 + S4

S = [ 100 + 412.5 + 2450 + 350]

S = 3312.5m

II) Average velocity = S / T

T = [ 10 + 15 + 70 + 20] s

T = 115s

V = 3312.5/115 = 28.8 m/s

Average velocity spent for the journey = 28.8m/s

III) Average Retardation = (v-u)/t4

V = 0

U = 35m/s

AR = (0 - 35)/20

AR = 1.75 m/ s (square)

Average Retardation as it's brought to rest = 1.75m/s^2

**Motion under gravity**

**Motion under the influence of gravity is based on acceleration due to gravity(g) of a free falling object and +/- g sign conventions.**

__Formula__

**Question 1**

**A cricket ball is thrown vertically upwards with an initial velocity of 40m/s . Find,**

I) it's velocity after 3s

II) Maximum height attained and the time it

III) The total time taken for the ball to return to the ground again.

Solution,

I) V = u - gt

V = 40 - 10 (3)

V = 40 -30 = 10m/s

II) using,

V = 0

U = 40m/s

g = 10m/s/s

From formula,

(40)(40) = 2 (10) h

1600 = 20 h

h =80 m

Therefore,

80 = 40t - 5t^2 ( it is now a quadratic equation)

16 = 8t - t^2

t^2 - 8t + 16 = 0

(t - 4)(t - 4) = 0

t = 4s

**Conclusion**

## 0 comments: