Question

An object has constant acceleration=4 m/s^2, it's velocity is= 1 at time=0 , when it is at distance=7 how fast is it moving when it is at distance= 8? At what time is it?

Answer 1

To solve this problem, we will use the equations of motion for an object with constant acceleration.

Given:
- Initial velocity \( v_0 = 1 \, \text{m/s} \) (at time \( t = 0 \))
- Acceleration \( a = 4 \, \text{m/s}^2 \)
- We want to find the velocity when the object is at a distance of \( 8 \, \text{m} \), and determine the time it takes to reach that distance.

### Step 1: Find the position function

We can use the following kinematic equation to find the position as a function of time:

\[
s(t) = s_0 + v_0 t + \frac{1}{2} a t^2
\]

Given that the object starts from \( s_0 = 0 \) at \( t = 0 \), the equation simplifies to:

\[
s(t) = v_0 t + \frac{1}{2} a t^2 = 1t + \frac{1}{2}(4)t^2 = t + 2t^2
\]

### Step 2: Set the position function equal to 8

Set \( s(t) = 8 \):

\[
t + 2t^2 = 8
\]

Rearranging gives:

\[
2t^2 + t - 8 = 0
\]

### Step 3: Solve the quadratic equation

Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Here, \( a = 2 \), \( b = 1 \), and \( c = -8 \):

\[
b^2 - 4ac = 1^2 - 4(2)(-8) = 1 + 64 = 65
\]

Now, applying the quadratic formula:

\[
t = \frac{-1 \pm \sqrt{65}}{4}
\]

Calculating the approximate values:

\[
t = \frac{-1 + \sqrt{65}}{4} \quad \text{(we take the positive root since time cannot be negative)}
\]

\[
t \approx \frac{-1 + 8.062}{4} \approx \frac{7.062}{4} \approx 1.7655 \, \text{s}
\]

### Step 4: Find the velocity when \( s = 8 \)

The velocity function is given by:

\[
v(t) = v_0 + at
\]
\[
v(t) = 1 + 4t
\]

Substituting \( t \approx 1.7655 \) into the velocity equation:

\[
v(1.7655) = 1 + 4(1.7655) \approx 1 + 7.062 \approx 8.062 \, \text{m/s}
\]

### Summary of Findings

- The speed of the object when it is at a distance of 8 m is approximately **8.06 m/s**.
- The time it takes to reach that position is approximately **1.77 seconds**.