Motion In A Straight Line

To find , we need to first understand the given relationship . Our aim is to express in terms of , and then differentiate with respect to .

First, solve for :

1. Rewrite the equation: .
2. Isolate by subtracting 4 from both sides: .
3. Square both sides to eliminate the square root: .

Now, differentiate with respect to :

1. Apply the chain rule to differentiate: .
2. Use the power rule: .
3. Since the derivative of with respect to is 1, we get: .

Evaluate at :

1. Substitute into the expression: .
2. Evaluate: .

Therefore, the value of is 0, which matches the correct answer.

[L]=

So, [v_²]²[a_²]=

[v₂]²[a₂]= or [L₂]= [L₁]

Similarly, [T]=

So, [T₂]= [T₁]

The correct option is (A): = and =

Let's consider the upward direction as positive.
The initial velocity of the particle, u = 150
Acceleration due to gravity, g = -10 (negative sign as it acts in the downward direction)
Using the kinematic equation, we can find the velocity of the particle at any time t as:
v = u - gt
At t = 3 s, the velocity of the particle is:
v₁ = u - gt
= 150 - 10(3)
= 120
At t = 5 s, the velocity of the particle is:
v₂ = u - gt
= 150 - 10(5)
= 100

Therefore, the ratio of the speed of the particle at t=3s to t=5s is:
= =
the ratio of the speed of the particle at t=3s to t=5s is 6:5. i.e 1.20

Solution:

Using the principle of energy conservation:

Where:

• is the initial kinetic energy,
• is the initial potential energy,
• is the final kinetic energy,
• is the final potential energy.

At the initial height, potential energy is:

At the final height, the body strikes the surface, and the potential energy is:

The body is released, meaning its initial velocity is zero. Therefore, .

Using conservation of energy:

Simplifying for velocity :

Substitute :

Final Answer:

The velocity with which the body strikes the surface is .

To solve this problem, we need to determine the distance between two cars when they come to rest after braking. Let's break it down step by step:

1. Both cars are traveling towards each other at a speed of .
2. The initial distance between the cars is .
3. The cars start applying brakes with a deceleration (negative acceleration) of .

We use the kinematic equation for constant acceleration:

• Here, is the final velocity (0 m/s, as the cars come to rest),
• is the initial velocity ( ),
• is the acceleration ( ),
• is the distance covered after applying brakes until the car stops.

Apply the equation separately for each car:

1. For one car:
   • Solving for gives:
2. Both cars travel each before stopping.

The combined distance covered by both cars is . Since the initial distance between the cars was , the distance remaining between the two cars when they stop is:

Therefore, the distance between the cars when they come to rest is 100 m.