A particle starts moving from rest from a fixed point in a fixed direction. The distance s from the fixed point at a time t is given by s = `t^(2)+at - b+ 17`, where a, b are real numbers. It the particle comes to rest after 5 sec at a distance of s = 25 units from the fixed point, then values of a and b are respectively

Options

A. 10,-33
B. `-10,-33`
C. `-8,33`
D. `-10,33`

Answer

To solve the problem step by step, we will follow the given conditions and equations.

### Step 1: Write down the given equation for distance
The distance \( s \) from a fixed point at time \( t \) is given by:
\[ s = t^2 + at - b + 17 \]

### Step 2: Find the expression for velocity
The velocity \( v \) of the particle is the derivative of the distance \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}(t^2 + at - b + 17) \]
Differentiating term by term, we get:
\[ v = 2t + a \]

### Step 3: Set the conditions for coming to rest
The particle comes to rest after 5 seconds, which means the velocity at \( t = 5 \) seconds is 0:
\[ v(5) = 0 \]
Substituting \( t = 5 \) into the velocity equation:
\[ 2(5) + a = 0 \]
This simplifies to:
\[ 10 + a = 0 \]
Thus, we find:
\[ a = -10 \]

### Step 4: Use the distance condition
We know that at \( t = 5 \) seconds, the distance \( s \) is 25 units:
\[ s(5) = 25 \]
Substituting \( t = 5 \) into the distance equation:
\[ s = (5^2) + a(5) - b + 17 \]
Substituting \( a = -10 \):
\[ 25 = 25 + (-10)(5) - b + 17 \]
This simplifies to:
\[ 25 = 25 - 50 - b + 17 \]
Combining like terms:
\[ 25 = -8 - b \]
Rearranging gives:
\[ b = -8 - 25 = -33 \]

### Step 5: Final values
Thus, we have:
\[ a = -10 \quad \text{and} \quad b = -33 \]

### Summary of the solution
The values of \( a \) and \( b \) are:
\[ \boxed{-10} \quad \text{and} \quad \boxed{-33} \]