Finding Acceleration and Distance with Constant Acceleration

Let's break down this physics problem involving a particle starting from rest with a constant acceleration. We are given information about the particle's speed at two different times and need to find the acceleration and the distance covered during a specific time interval.

Understanding the Problem Setup

The particle begins its motion from rest, meaning its initial velocity ($u$) is 0 m/s. It accelerates at a constant rate, let's call it $a$. We are given the speed at time $t$ and the speed one second later at time $(t+1)$.

• Initial velocity, $u = 0$ m/s
• Speed at time $t$, $v_t = 100$ m/s
• Speed at time $(t+1)$, $v_{t+1} = 150$ m/s
• Acceleration, $a$ = constant

We need to find the value of the constant acceleration, $a$, and the distance the particle travels specifically during the $(t+1)$th second.

Applying Kinematic Equations

We can use the kinematic equation relating velocity, initial velocity, acceleration, and time:

$v = u + at$

Using the given information:

1. At time $t$, the speed is 100 m/s:

   \( v_t = u + a \cdot t \)

   \( 100 = 0 + a \cdot t \)

   \( 100 = at \)   (Equation 1)

2. At time $(t+1)$, the speed is 150 m/s:

   \( v_{t+1} = u + a \cdot (t+1) \)

   \( 150 = 0 + a(t+1) \)

   \( 150 = at + a \)    (Equation 2)

Calculating the Acceleration

Now we have two equations with two unknowns, $a$ and $t$. We can solve these equations simultaneously. Substitute the value of $at$ from Equation 1 into Equation 2:

\( 150 = (at) + a \)

\( 150 = 100 + a \)

Subtract 100 from both sides:

\( a = 150 - 100 \)

\( a = 50 \text{ m/s}^2 \)

The acceleration of the particle is 50 m/s\(\text{ m/s}^2\).

Finding the Time 't'

Now that we know the acceleration, we can find the value of $t$ using Equation 1:

\( 100 = at \)

\( 100 = 50 \cdot t \)

Divide by 50:

\( t = \frac{100}{50} \)

\( t = 2 \text{ seconds} \)

So, the speed was 100 m/s at $t=2$ seconds, and 150 m/s at $t+1 = 3$ seconds.

Calculating Distance in the (t+1)th Second

We need to find the distance travelled during the $(t+1)$th second, which is the 3rd second in this case ($t+1 = 2+1=3$). The formula for the distance travelled in the $n$th second is:

\( S_n = u + \frac{a}{2}(2n - 1) \)

Here, $u=0$ (starting from rest), $a=50$ m/s\(\text{ m/s}^2\), and $n = t+1 = 3$.

Substitute these values into the formula:

\( S_{3} = 0 + \frac{50}{2}(2 \cdot 3 - 1) \)

\( S_{3} = 25(6 - 1) \)

\( S_{3} = 25(5) \)

\( S_{3} = 125 \text{ meters} \)

The distance travelled during the (t+1)th (or 3rd) second is 125 meters.

Final Result Summary

The value of the constant acceleration is 50 m/s\(\text{ m/s}^2\), and the distance travelled during the (t+1)th second is 125 m.