A particle moves on a line according to the law `s=at^(2)+bt+c`. If the displacement after 1 sec is 16 cm, the velocity after 2 sec is 24 cm/sec and acceleration is `8cm//sec^(2)`, then

Options

A. a=4, b=8, c=4
B. a=4, b=4, c=8
C. a=8, b=4, c=4
D. none of these

Answer

To solve the problem, we have the displacement of a particle given by the equation:

\[ s = at^2 + bt + c \]

where \( s \) is the displacement, \( t \) is the time, and \( a \), \( b \), and \( c \) are constants we need to determine.

### Step 1: Find the acceleration
The acceleration \( a \) is given as \( 8 \, \text{cm/s}^2 \). The acceleration can be derived from the displacement function by differentiating twice:

1. First, we find the velocity \( v \) by differentiating \( s \):
\[ v = \frac{ds}{dt} = 2at + b \]

2. Then, we find the acceleration \( a \) by differentiating the velocity:
\[ a = \frac{dv}{dt} = 2a \]

Given that the acceleration is \( 8 \, \text{cm/s}^2 \), we can set up the equation:
\[ 2a = 8 \]
From this, we can solve for \( a \):
\[ a = 4 \]

### Step 2: Find the velocity at \( t = 2 \) seconds
We know the velocity at \( t = 2 \) seconds is \( 24 \, \text{cm/s} \). Using the velocity equation we derived:
\[ v = 2at + b \]
Substituting \( a = 4 \) and \( t = 2 \):
\[ v = 2(4)(2) + b = 16 + b \]
Setting this equal to the given velocity:
\[ 16 + b = 24 \]
Solving for \( b \):
\[ b = 24 - 16 = 8 \]

### Step 3: Find the displacement at \( t = 1 \) second
We are given that the displacement at \( t = 1 \) second is \( 16 \, \text{cm} \). Using the displacement equation:
\[ s = at^2 + bt + c \]
Substituting \( t = 1 \):
\[ s = 4(1^2) + 8(1) + c = 4 + 8 + c = 12 + c \]
Setting this equal to the given displacement:
\[ 12 + c = 16 \]
Solving for \( c \):
\[ c = 16 - 12 = 4 \]

### Final Values
We have found:
- \( a = 4 \)
- \( b = 8 \)
- \( c = 4 \)

Thus, the values of \( a \), \( b \), and \( c \) are \( 4, 8, 4 \) respectively.

### Summary
The values of \( a, b, c \) are:
- \( a = 4 \)
- \( b = 8 \)
- \( c = 4 \)