the acceleration of a particle is given by a=3-2t

(a)find the initial speed v0 such that the particle will have the same x-coordinate at t=5s as it had at t=0.

(b)what will be the velocity at t=5s?

To tackle the problem of a particle whose acceleration is defined by the equation \( a = 3 - 2t \), we need to find both the initial speed \( v_0 \) and the velocity at \( t = 5 \) seconds. Let's break this down step by step.

Step 1: Understanding Acceleration and Velocity

Acceleration is the rate of change of velocity with respect to time. In this case, we have a time-dependent acceleration. To find the velocity, we need to integrate the acceleration function over time.

Finding Velocity as a Function of Time

The acceleration function is given as:

a(t) = 3 - 2t

To find the velocity, we integrate the acceleration with respect to time:

v(t) = ∫a(t) dt = ∫(3 - 2t) dt

Carrying out the integration gives:

v(t) = 3t - t^2 + C

Here, \( C \) is the constant of integration, which represents the initial velocity \( v_0 \) at \( t = 0 \).

Step 2: Finding Position as a Function of Time

Next, we need to find the position function \( x(t) \) by integrating the velocity function:

x(t) = ∫v(t) dt = ∫(3t - t^2 + C) dt

Integrating this gives:

x(t) = (3/2)t^2 - (1/3)t^3 + Ct + D

Here, \( D \) is another constant of integration representing the initial position \( x_0 \) at \( t = 0 \).

Step 3: Applying the Condition for Same x-coordinate

We want the particle to have the same x-coordinate at \( t = 5 \) seconds as it had at \( t = 0 \). This means:

x(5) = x(0)

Calculating \( x(0) \):

x(0) = D

Calculating \( x(5) \):

x(5) = (3/2)(5^2) - (1/3)(5^3) + 5C + D

Substituting the values:

x(5) = (3/2)(25) - (1/3)(125) + 5C + D

x(5) = 37.5 - 41.67 + 5C + D

Setting \( x(5) = x(0) \) gives:

37.5 - 41.67 + 5C + D = D

This simplifies to:

5C - 4.17 = 0

From this, we can solve for \( C \):

C = 0.834

Thus, the initial speed \( v_0 \) is approximately \( 0.834 \, \text{m/s} \).

Step 4: Calculating Velocity at t = 5s

Now that we have \( v_0 \), we can find the velocity at \( t = 5 \) seconds:

v(5) = 3(5) - (5^2) + C

Substituting the values:

v(5) = 15 - 25 + 0.834

v(5) = -9.166 \, \text{m/s}

Summary of Results

• Initial speed \( v_0 \approx 0.834 \, \text{m/s} \)
• Velocity at \( t = 5 \, \text{s} \) is \( v(5) \approx -9.166 \, \text{m/s} \)

This analysis shows how to derive velocity and position from acceleration, and how to apply conditions to find specific values. If you have any further questions or need clarification on any part, feel free to ask!