Test: Rectilinear Motion - Civil Engineering (CE) MCQ

Given,  t = αx² + βx
Differentiating the above equation with respect to x,
dt/dx = 2ax + b ...(1)
We, know that the rate of change of displacement with respect to time is velocity, V = dt/dx
From equation 1.

The rate of change of velocity with respect to time is acceleration or retardation(a),
Differentiating equation 2 with respect to time,

From equation 2 and 3.
a = -2αV³
Hence, Retardation = 2αV³

Calculation:
Given:
Velocity, V = (3t² + 2t) m/s, Position = ?, at t = 3 sec

So,
Position(x) = t³ + t²
putting t = 3 sec.,
(x)_t=3 = 36 m

Acceleration:

• The object under motion can undergo a change in its speed. The measure of the rate of change in its speed along with direction with respect to time is called acceleration.
• The motion of the object can be linear or circular.

Linear acceleration:

• The acceleration involved in linear motion is called linear acceleration.
• Here the acceleration is only due to the change in speed and no acceleration due to change in direction, therefore no radial component of acceleration.

Circular acceleration:

• The acceleration involved in a circular motion is called angular acceleration.
• In a circular motion, the acceleration experienced by the body towards the centre is called the centripetal acceleration which can be resolved into two-component.
• A radial component and a tangential component depending upon the type of motion.

Radial acceleration (a_r):

• The acceleration of the object along the radius, directed towards the centre is called radial acceleration.
  

Tangential acceleration (a_t):

• The tangential component is defined as the component of angular acceleration tangential to the circular path.
  

We know, 
and u = 0

Hence 
Apparently, the ratio of times taken is independent of the mass of bodies.

Acceleration:

• The object under motion can undergo a change in its speed. The measure of the rate of change in its speed along with direction with respect to time is called acceleration.
• The motion of the object can be linear or circular.

Linear acceleration:

• The acceleration involved in linear motion is called linear acceleration.
• Here the acceleration is only due to the change in speed and no acceleration due to change in direction, therefore no radial component of acceleration.

Circular acceleration:

• The acceleration involved in a circular motion is called angular acceleration.
• In a circular motion, the acceleration experienced by the body towards the centre is called the centripetal acceleration which can be resolved into two-component.
• A radial component and a tangential component depending upon the type of motion.

Radial acceleration (a_r):

• The acceleration of the object along the radius, directed towards the centre is called radial acceleration.
  

Tangential acceleration (a_t):

• The tangential component is defined as the component of angular acceleration tangential to the circular path.
  

Acceleration:

• The object under motion can undergo a change in its speed. The measure of the rate of change in its speed along with direction with respect to time is called acceleration.
• The motion of the object can be linear or circular.

Linear acceleration:

• The acceleration involved in linear motion is called linear acceleration.
• Here the acceleration is only due to the change in speed and no acceleration due to change in direction, therefore no radial component of acceleration.

Circular acceleration:

• The acceleration involved in a circular motion is called angular acceleration.
• In a circular motion, the acceleration experienced by the body towards the centre is called the centripetal acceleration which can be resolved into two-component.
• A radial component and a tangential component depending upon the type of motion.

Radial acceleration (a_r):

• The acceleration of the object along the radius, directed towards the centre is called radial acceleration.
  

Tangential acceleration (a_t):

• The tangential component is defined as the component of angular acceleration tangential to the circular path.
  

Basic condition of a rigid body is,

Considering two points P and Q on a rigid body.
The relative position of point P with respect to Q is r_PQ. 
The relative velocity of P with respect to Q is V_PQ = ωṙ_PQ .
If the relative distance between the P and Q does not change then V_PQ = ωṙ_PQ .
V_PQ = ω × ṙP/Q 
ω is the angular velocity of the rigid body.
Since V_PQ is the cross product of ω and ṙ_P/Q the relative velocity of P with respect to Q will be perpendicular to PQ.

According to the laws of motion, the distance covered by a body is given as, s = ut + (1/2)at²
Calculation:
Given:
s = 450 m in t = 5 second and further s = 700 m in t = 10 second
where s = distance travelled and t = time taken for both case saperately.
When car covers 450 m

Now,
When the car covers the remaining 700 m.

Solving equaiton (1) and (2)
u = 65 m/sec
∴ a = 10 m/sec²

Under the given situation the ball will go up reach to a maximum height then come down, with the same initial velocity it will hit the ground but when it bounced back it will loss 20% of the velocity hence will move up with an initial velocity of 80% of the previous initial velocity. Thus, the time of flight will keep on decreasing with the collisions. The situation will look like this.
Let (v_i) be the initial velocity, after (i^th) collision
From the data we know (Loss of velocity 20%)
(u₁ = 4 m/s) * and [u_(i+1) = 0.08 U_i]
Now the time of flight in each collision: [time to go up = time to come down]
v = u + at
Where, a ⇒ (g) = -10 m/s² ↑ (- ve acceleration)
u = Initial velocity; v = final velocity
At the top (v = 0); t = time taken
0 = u – gt ⇒ t = (u/g)
Now, total time of flight = (2 times) the time to go up. 
So, in (i^th) collision the time of flight is 
Total time will be summation of all the time of flight

Now, this looks like a geometric progression (G.P) with ratio of (0.8)

Now to final the sum of the G.P, [1 + 0.8 + (0.8)² + (0.8)³ + ….]



Mistakes
Here, the mistake that can happen is one can confuse 20% loss as (20% of the velocity). So, instead (0.8 u) You will take (0.2 u) as the velocity after the collision which will result in following.

Which is not correct
Another mistake is the time of flight calculation
  instead you might commit a mistake of taking (u/g) as the time of flight in hurry as you forget to take the time of ball to come down which is same as time to go up.
Which will result in (1/2) of the answer
⇒ (1/2) x 4 = 2 sec which is not correct.

Given,  t = αx² + βx
Differentiating the above equation with respect to x,

We, know that the rate of change of displacement with respect to time is velocity, V = dx/dt
From equation 1.

The rate of change of velocity with respect to time is acceleration or retardation(a),
Differentiating equation 2 with respect to time,

From equation 2 and 3.
a = -2αV³
Hence, Retardation = 2αV³.