For a particle undergoing rectilinear motion with uniform acceleration, the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude of initial velocity for this time interval. Then the ratio of initial speed to the final speed for this thime interval is:

Question

For a particle undergoing rectilinear motion with uniform acceleration, the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude of initial velocity for this time interval. Then the ratio of initial speed to the final speed for this thime interval is:

Answer 1

This ans is wrong

Answer 2

Let's denote the magnitude of initial velocity as "u" and the magnitude of final velocity as "v". We are given two conditions:

1. The magnitude of displacement is one third the distance covered in the time interval.

From the equation of motion, for rectilinear motion with uniform acceleration, we have:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

Since the particle undergoes uniform acceleration, we can use the following equations:

s = ut + 0.5at^2 (1)
v = u + at (2)

From condition 1, the magnitude of displacement (s) is 1/3 times the distance covered. Let's denote the distance covered as "d". Therefore, we have:

s = (1/3)d

From equation (1), we can express "d" in terms of "u" and "t":
d = ut + 0.5at^2

Substituting this value of "d" in the equation for displacement (s), we get:

s = (1/3)(ut + 0.5at^2)

Now, let's consider the second condition:

2. The magnitude of final velocity (v) is less than the magnitude of initial velocity (u).

From equation (2), we have:

v = u + at

Since v < u, we can rewrite it as:

0 < u - v

Now, let's find the ratio of initial speed (magnitude of initial velocity) to the final speed(magnitude of final velocity).

Ratio = u/v

Substituting the value of v from equation (2), we get:

Ratio = u / (u + at)

We can further simplify this expression by multiplying the numerator and the denominator by t:

Ratio = (u * t) / (u * t + at^2)

Since we have an expression for s in terms of u and t, we can substitute it in the denominator:

Ratio = (u * t) / ((1/3)(ut + 0.5at^2) + at^2)

Simplifying the denominator:

Ratio = (u * t) / ((1/3)(ut + 0.5at^2) + (3/3)at^2)

Ratio = (u * t) / ((1/3)(ut + 0.5at^2) + (3/3)(at^2))

Ratio = (u * t) / ((ut + 0.5at^2) / 3 + (3at^2) / 3)

Simplifying further:

Ratio = (u * t) / ((ut + 0.5at^2 + 3at^2) / 3)

Ratio = 3u * t / (ut + 3.5at^2)

So, the ratio of initial speed to the final speed for this time interval is 3u * t / (ut + 3.5at^2).

Answer 3

To find the ratio of initial speed to the final speed for the given time interval, we need to apply the equations of motion.

Let's assume:
Initial velocity = u
Final velocity = v
Time interval = t
Acceleration = a
Distance covered = s

We are given two conditions:

1) The magnitude of displacement is one third the distance covered in some time interval.
This can be written as:
|s| = (1/3)|s|

Since magnitude cannot be negative, we can remove the absolute value signs:
s = (1/3)s

2) The magnitude of the final velocity is less than the magnitude of the initial velocity.
|v| < |u|

Now, let's use the equations of motion to establish the relationship between the given quantities.

Using the equation of motion relating displacement, initial velocity, time, and acceleration:
s = ut + (1/2)at^2

Since the question mentions uniform acceleration, we can simplify this equation to:
s = ut + (1/2)at^2

Substituting the condition s = (1/3)s, we get:
(1/3)s = ut + (1/2)at^2

Now, let's find the expression for final velocity (v) using the equation of motion:
v = u + at

Substituting the condition |v| < |u|:
|u + at| < |u|

Since magnitude cannot be negative, we have two cases:
1) u + at < u
2) -(u + at) < u

Simplifying both cases separately, we get:
1) at < 0 (which is not possible, as time cannot be negative)
2) at > -2u

Now, let's find the ratio of initial speed (|u|) to the final speed (|v|) using the expressions we derived.

Ratio = |u|/|v|
Since |v| < |u|, the ratio is greater than 1.

Therefore, the ratio of initial speed to the final speed for this time interval is greater than 1.

Answer 4

it evidently goes forward a distance 3d then back a distance 2d to end up a displacoutup to be negative.

deceleration phase out to 3 d
t1 = time out to 3 d
v = Vi + a t1
0 = Vi + a t1
so
t1 = -Vi/a
3 d = Vi t1 + (1/2) a t1^2
3d = -Vi^2/a + .5 Vi^2/a = -.5Vi^2/a
d a = -(1/6) Vi^2

return phase back from 3 d to one d
Vf = 0 + a t2
so
t2 = Vf/a
-2d = (1/2) a t2^2
-2d = .5 Vf^2/a
--------------------------------
algebra from here on :)
d a = -(1/4) Vf^2
(1/6) Vi^2 = (1/4)Vf^2