A particle starts from rest, moving in a straight line with a time varying acceleration `( "in"m//s^(2) ): a= 30 -(15)/(2) sqrt(t) ,` where time is in seconds . The displacement of the particle in 4 seconds is .

Explanation

To solve the problem step by step, we will follow the process of integrating the acceleration function to find the velocity and then the displacement.

### Step 1: Write down the acceleration function
The acceleration of the particle is given by:
\[ a(t) = 30 - \frac{15}{2} \sqrt{t} \]

### Step 2: Relate acceleration to velocity
Acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = 30 - \frac{15}{2} \sqrt{t} \]

### Step 3: Integrate to find velocity
To find the velocity, we integrate both sides with respect to time \(t\):
\[ dv = \left(30 - \frac{15}{2} \sqrt{t}\right) dt \]
Integrating both sides:
\[ v(t) = \int \left(30 - \frac{15}{2} \sqrt{t}\right) dt \]
Calculating the integral:
\[ v(t) = 30t - \frac{15}{2} \cdot \frac{2}{3} t^{3/2} + C \]
\[ v(t) = 30t - 5t^{3/2} + C \]
Since the particle starts from rest, when \(t = 0\), \(v(0) = 0\):
\[ 0 = 30(0) - 5(0)^{3/2} + C \implies C = 0 \]
Thus, the velocity function is:
\[ v(t) = 30t - 5t^{3/2} \]

### Step 4: Relate velocity to displacement
Velocity is also the derivative of displacement with respect to time:
\[ v = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = 30t - 5t^{3/2} \]

### Step 5: Integrate to find displacement
To find the displacement, we integrate the velocity function:
\[ dx = (30t - 5t^{3/2}) dt \]
Integrating both sides:
\[ x(t) = \int (30t - 5t^{3/2}) dt \]
Calculating the integral:
\[ x(t) = 15t^2 - \frac{5}{\frac{5}{2}} t^{5/2} + C \]
\[ x(t) = 15t^2 - 2t^{5/2} + C \]
Since the particle starts from the origin, when \(t = 0\), \(x(0) = 0\):
\[ 0 = 15(0)^2 - 2(0)^{5/2} + C \implies C = 0 \]
Thus, the displacement function is:
\[ x(t) = 15t^2 - 2t^{5/2} \]

### Step 6: Calculate displacement at \(t = 4\) seconds
Now, we find the displacement at \(t = 4\):
\[ x(4) = 15(4^2) - 2(4^{5/2}) \]
Calculating each term:
\[ x(4) = 15(16) - 2(32) = 240 - 64 = 176 \]

### Final Answer
The displacement of the particle in 4 seconds is:
\[ \boxed{176 \text{ meters}} \]

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