IEMDaily - Video Lecture Notes

Example

Missing Variables in Problem involving second equation of motion

A car has initial speed

3 m / s

and it takes

1 s

to travel

10 m

then what will be the acceleration of car?
Since

s = u t + 1 / 2 a t^{2}

10 = 3 \times 1 + 1 / 2 a \times 1^{2}

a = 14 m / s^{2}

Example

Finding Missing Variables of Third Equation of Motion

If a particle is moving with velocity

5 m / s

. And it has an acceleration of

2 m / s^{2}

. After a while its velocity reaches to

10 m / s

. Then what will be the distance traveled by this particle?

v^{2} = u^{2} + 2 a s

10^{2} = 5^{2} + 2 \times 2 \times s

s = 18.75 m

Example

Problems Involving more than one Equation of Motion

Example: A fly flying at a velocity of 5m/s and it speeds up for 1s and its velocity reaches to

9 m / s

. What will be the distance covered by the fly during this motion?

Solution:

v = u + a t

9 = 5 + a \times 1^{2}

a = 4 m / s^{2}

Now

9^{2} = 5^{2} + 2 \times 4 \times s

s = 7 m

Example

Total distance traveled when velocity and accelerations are in opposite directions

If a particle has initial velocity as 10m/s and it's retarding at a rate of

- 1 m / s^{2}

then distance travel by this in one second is:

s = u t + 1 / 2 a t^{2}

s = 10 \times 1 - 1 / 2 \times 1 \times 1^{2}

s = 9.5 m

Example

Motion in different acceleration for different time intervals

A particle started its motion from rest with an acceleration of

1 m / s^{2}

for 2s and then continued it for next 1s with an acceleration of

2 m / s^{2}

. The distance traveled during this will be:
After 2s final velocity is:

v = u + a t

v = 0 + 1 \times 2

v = 2 m / s

Now this is the initial velocity for the second half of the motion.

s = u t + 1 / 2 a t^{2}

s = 2 \times 1 + 1 / 2 \times 2 \times 1^{2}

s = 3 m

Distance traveled in first half is:

s = 1 / 2 a t^{2}

s = 1 / 2 \times 1 \times 2^{2}

s = 2 m

Hence total distance traveled

= 3 + 2 = 5 m

Example

Motion involving more than one uniform acceleration

Example: A particle is moving along x-axis with zero initial velocity for

2 s

with an acceleration of

1 m / s^{2}

. After this it takes a

90

degree turn and starts moving along Y-axis with same acceleration. What will be distance covered by this particle along Y-direction after

4 s

of motion.

Solution:
Final Velocity along x-axis is:
from

v = u + a t

v = 2 \times 1 = 2 m / s

s = u t + 1 / 2 a t^{2}

s = 2 \times 2 + 1 / 2 \times 1 \times 2^{2}

s = 6 m

Example

problem on acceleration as a function of velocity

Problem: A point moves rectilinearly with deceleration whose modulus depends on the velocity

v

of the particle as

ω = - a \sqrt{v}

, where

a

is a positive constant. At the initial moment the velocity of the point is equal to

v_{0}

.What distance will it traverse before it stops?
Solution:

ω = - a \sqrt{v},

minus sign for deceleration

v \frac{d v}{d x} = - a \sqrt{v}

Integrating,

\int_{v_{0}}^{0} v^{1 / 2} d v = - a \int_{0}^{x_{0}} d x x_{0} = \frac{2 {v_{0}}^{3 / 2}}{3 a}

Example

problem on acceleration as a function of displacement

Problem: The acceleration,

a

, of a particle depends on displacement,

s

, as

a = s + 5

. It is given that initially

s = 0

and

v = 5 m / s

. Then, the expression for velocity

v

as a function of

s

.

Solution:Here,

a = s + 5

So,

\frac{d v}{d t} = s + 5

\frac{d v}{d s} \frac{d s}{d t} = s + 5

\frac{d v}{d s} v = s + 5

(as

\frac{d s}{d t} = v

)or

v d v = (s + 5) d s

Integrating both sides,

\int v d v = \int (s + 5) d s

\frac{v^{2}}{2} = \frac{s^{2}}{2} + 5 s + C

where C is integrating constant
When

S = 0, v = 5, C = 25 / 2

So,

v^{2} / 2 = s^{2} / 2 + 5 s + 25 / 2

v^{2} = s^{2} + 10 s + 25

v^{2} = (s + 5)^{2}

v = s + 5

Definition

Relative Displacement

Relative displacement, which is displacement of a point on a structure with respect to its original location or an adjacent point on the structure that has also undergone movement, can be an effective indicator of post event structural damage.

Relative displacement is

\vec{r} = {\vec{r}}_{1} - {\vec{r}}_{2}

Definition

Motion of a body

A body is said to be in a state of rest w.r.t. a stationary body if its position does not change with time as seen by the stationary body. Similarly, it is in the state of motion w.r.t a stationary body if its position changes with time as seen by the stationary body.