Let's solve the problem step by step.

Given:

1. Distance covered at the 11th second, $s_{11} = 720$ cm
2. Distance covered at the 15th second, $s_{15} = 96$ cm
3. The particle is moving with a constant deceleration.

Required:

1. Initial velocity $u$
2. Deceleration $a$

Using the equation of motion:

The position of the particle at any time $t$ is given by:

$s_t = ut - \frac{1}{2}at^2$

For the 11th second (t = 11):

$s_{11} = u(11) - \frac{1}{2}a(11)^2$

$720 = 11u - \frac{121}{2}a \quad \text{(Equation 1)}$

For the 15th second (t = 15):

$s_{15} = u(15) - \frac{1}{2}a(15)^2$

$96 = 15u - \frac{225}{2}a \quad \text{(Equation 2)}$

Solving the equations:

From Equation 1: $720 = 11u - 60.5a \quad \text{(1)}$

From Equation 2: $96 = 15u - 112.5a \quad \text{(2)}$

Now, subtract Equation 1 from Equation 2 to eliminate $u$:

$(96 - 720) = (15u - 11u) - (112.5a - 60.5a)$

$-624 = 4u - 52a$

$4u - 52a = -624 \quad \text{(Equation 3)}$

Simplify Equation 3:

$u - 13a = -156 \quad \text{(Equation 4)}$

Now, let's substitute Equation 4 into Equation 1 to find $a$:

From Equation 1: $720 = 11(-156 + 13a) - 60.5a$

$720 = -1716 + 143a - 60.5a$

$720 + 1716 = 82.5a$

$2436 = 82.5a$

$a = \frac{2436}{82.5} \approx 29.52 \text{ cm/s}^2$

Now, substitute $a$ back into Equation 4 to find $u$:

$u = -156 + 13(29.52)$

$u = -156 + 383.76 \approx 227.76 \text{ cm/s}$

Final Answers:

• Initial Velocity $u \approx 227.76$ cm/s
• Deceleration $a \approx 29.52$ cm/s²

Would you like further details or explanations?

Here are some related questions to explore further:

1. How does constant deceleration affect the motion of an object?
2. How would the initial velocity change if the time intervals were different?
3. What would be the effect on the motion if the deceleration were not constant?
4. How can we derive the equation of motion for a particle with constant deceleration?
5. What are the implications of a negative initial velocity in this context?

Tip: When solving problems involving constant acceleration or deceleration, always carefully apply the correct kinematic equations to ensure accuracy.