Question

An object starts from rest and accelerates uniformly for 3.5 seconds. If its final velocity is 25 m/s, what is its displacement?

Answer 1

To find the displacement of the object, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. Since the object starts from rest, the initial velocity \( u = 0 \, \text{m/s} \), the final velocity \( v = 25 \, \text{m/s} \), and the time \( t = 3.5 \, \text{s} \).

First, we need to calculate the acceleration \( a \) using the formula:

\[
v = u + at
\]

Rearranging to solve for \( a \):

\[
a = \frac{v - u}{t} = \frac{25 \, \text{m/s} - 0 \, \text{m/s}}{3.5 \, \text{s}} = \frac{25}{3.5} \approx 7.14 \, \text{m/s}^2
\]

Now that we have the acceleration, we can calculate the displacement \( s \) using the following kinematic equation:

\[
s = ut + \frac{1}{2} a t^2
\]

Substituting in the known values (with \( u = 0 \, \text{m/s} \)):

\[
s = 0 + \frac{1}{2} a t^2 = \frac{1}{2} \cdot (7.14 \, \text{m/s}^2) \cdot (3.5 \, \text{s})^2
\]

Calculating \( t^2 \):

\[
(3.5 \, \text{s})^2 = 12.25 \, \text{s}^2
\]

Now substituting this back in:

\[
s = \frac{1}{2} \cdot 7.14 \, \text{m/s}^2 \cdot 12.25 \, \text{s}^2 = \frac{1}{2} \cdot 87.625 \, \text{m} \approx 43.81 \, \text{m}
\]

Therefore, the displacement of the object is approximately **43.81 meters**.