Question Statement

A particle $P$ moves along the $x$-axis. The acceleration of $P$ at time $t$ seconds, when $t\ge 0$, is $a=(3t+5){\text{ m/s}}^2$ in the positive $x$-direction. When $t=0$, the velocity of $P$ is $2\text{ m/s}$ in the positive $x$-direction. When $t=T$, the velocity of $P$ is $6\text{ m/s}$. Find the value of $T$.

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Background and Explanation

This problem involves kinematics in one dimension, specifically the relationship between acceleration and velocity. Since acceleration is the rate of change of velocity with respect to time, you can find the velocity function by integrating the acceleration function and applying the given initial conditions to determine the constant of integration.

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Solution

We are given the acceleration function:

$a=3t+5{\text{m/s}}^2$

With boundary conditions:

• At $t=0$, $v=2$ m/s
• At $t=T$, $v=6$ m/s

Finding the Velocity Function

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

$v=\int adt=\int (3t+5)dt$

Using the power rule for integration:

$v=\frac{3}{2}{t}^2+5t+C$

where $C$ is the constant of integration.

Applying the Initial Condition

Using the initial condition that at $t=0$, $v=2$:

$2=\frac{3}{2}(0{)}^2+5(0)+C$

$2=C$

Therefore, the velocity function is:

$v=\frac{3}{2}{t}^2+5t+2$

Finding $T$ Using the Final Condition

At $t=T$, the velocity is $6$ m/s. Substituting into our velocity equation:

$6=\frac{3}{2}{T}^2+5T+2$

Rearranging to form a quadratic equation:

$\frac{3}{2}{T}^2+5T+2-6=0$

$\frac{3}{2}{T}^2+5T-4=0$

To eliminate the fraction, multiply the entire equation by $2$:

$3T^2+10T-8=0$

Solving the Quadratic Equation

We solve by factorization. Looking for two numbers that multiply to $3\times (-8)=-24$ and add to $10$. These numbers are $12$ and $-2$.

$3T^2+12T-2T-8=0$

Factor by grouping:

$3T(T+4)-2(T+4)=0$

$(T+4)(3T-2)=0$

This gives two possible solutions:

$T+4=0\Rightarrow T=-4$

or

$3T-2=0\Rightarrow 3T=2\Rightarrow T=\frac{2}{3}$

Since time $t\ge 0$ in this physical context, we reject the negative solution $T=-4$.

Therefore:

$T=\frac{2}{3}\text{ seconds}$

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Key Formulas or Methods Used

• Relationship between acceleration and velocity: $v=\int adt$ or $a=\frac{dv}{dt}$
• Power rule for integration: $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Solving quadratic equations: Factorization or quadratic formula $T=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Physical constraints: Time $t\ge 0$ in kinematics problems, allowing rejection of non-physical negative time solutions

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Summary of Steps

1. Write down the given acceleration: $a=3t+5$
2. Integrate to find velocity: $v=\frac{3}{2}{t}^2+5t+C$
3. Apply initial condition $(t=0,v=2)$ to find $C=2$
4. Write complete velocity equation: $v=\frac{3}{2}{t}^2+5t+2$
5. Substitute final condition $(t=T,v=6)$ to get $\frac{3}{2}{T}^2+5T+2=6$
6. Simplify to standard quadratic form: $3T^2+10T-8=0$
7. Factor and solve: $(T+4)(3T-2)=0$ yielding $T=-4$ or $T=\frac{2}{3}$
8. Select physical solution: $T=\frac{2}{3}$ s (rejecting negative time)