A PARTICLE IS MOVING IN A PLANE SUCH THAT ITS VELOCITY AT TIME T= 0 SECOND IS 56I^ + 45J^ + 90K^.

ALSO ITS VELOCITY IS RELATED TO THE TIME AS V = del/del(t)(3t^2 + 72t^56 +85z) + ln(2t+5). ..........(here del/del(t) is partial derivative with respect to t)

now, also its displacement varies with time as - integral (1/5.2t^2 + 78|ln(6t+5)| + v (integral means the integration symbol)

find the acceleration and displacement at time t = 60 seconds.

To solve the problem of finding the acceleration and displacement of the particle at time \( t = 60 \) seconds, we need to break down the information given and apply the appropriate mathematical concepts. Let's go through this step by step.

Understanding the Velocity Function

We start with the velocity of the particle at \( t = 0 \) seconds, which is given as:

V(0) = 56i + 45j + 90k

Next, we have a velocity function that is defined in terms of time:

V(t) = ∂/∂t (3t² + 72t⁵ + 85z) + ln(2t + 5)

To find the acceleration, we first need to compute the velocity function by taking the partial derivative of the expression with respect to \( t \).

Calculating the Partial Derivative

Let’s differentiate the components of the velocity function:

• For \( 3t² \), the derivative is \( 6t \).
• For \( 72t⁵ \), the derivative is \( 360t⁴ \).
• For \( 85z \), since it does not depend on \( t \), the derivative is \( 0 \).
• For \( ln(2t + 5) \), using the chain rule, the derivative is \( \frac{2}{2t + 5} \).

Putting it all together, we have:

V(t) = (6t + 360t⁴) + \frac{2}{2t + 5}

Finding the Acceleration

Acceleration is the derivative of velocity with respect to time:

A(t) = ∂V/∂t

Now, we differentiate \( V(t) \):

• The derivative of \( 6t \) is \( 6 \).
• The derivative of \( 360t⁴ \) is \( 1440t³ \).
• For \( \frac{2}{2t + 5} \), we apply the quotient rule, yielding \( -\frac{4}{(2t + 5)²} \).

Thus, the acceleration function becomes:

A(t) = 6 + 1440t³ - \frac{4}{(2t + 5)²}

Calculating Acceleration at \( t = 60 \) seconds

Now we substitute \( t = 60 \) into the acceleration function:

A(60) = 6 + 1440(60)³ - \frac{4}{(2(60) + 5)²}

Calculating each term:

• First term: \( 6 \).
• Second term: \( 1440 \times 216000 = 311040000 \).
• Third term: \( 2(60) + 5 = 125 \), so \( (125)² = 15625 \) and \( -\frac{4}{15625} \approx -0.000256 \).

Putting it all together:

A(60) ≈ 311040000 + 6 - 0.000256 ≈ 311040006

Calculating Displacement

The displacement is given by the integral of the velocity function:

S(t) = -∫(1/(5.2t² + 78|ln(6t + 5)| + V(t)) dt

To find the displacement at \( t = 60 \), we need to evaluate this integral. However, this integral may be complex and could require numerical methods or specific techniques to solve accurately. For simplicity, let’s denote the integral as:

S(60) = -I(60)

Where \( I(t) \) represents the integral of the expression. The exact evaluation of this integral may depend on the context or tools available, such as numerical integration techniques.

Final Results

In summary, we found:

• Acceleration at \( t = 60 \) seconds: A(60) ≈ 311040006
• Displacement at \( t = 60 \) seconds: S(60) = -I(60) (exact value depends on the integral evaluation).

Understanding these concepts and calculations is crucial in physics, especially in kinematics, where velocity, acceleration, and displacement are fundamental to describing motion.