Sure, let's tackle this problem step by step, using kinematic equations from physics to find the velocity and the time required.
### Step 1: Understand Initial Conditions and Equations
1. Initial Conditions:
- The car starts from rest, so its initial velocity \(\( u \)\) is 0 m/s.
- The acceleration \(\( a \)\) is 2 m/s².
- The distance \(\( s \)\) covered is 49 meters.
2. Equations Used:
- To find the final velocity (\(\( v \)\)), we use the kinematic equation:
\(\[
v^2 = u^2 + 2as
\]\)
- To find the time (\(\( t \)\)), we use the kinematic equation:
\(\[
s = ut + \frac{1}{2}at^2
\]\)
### Step 2: Calculate Final Velocity
The car starts from rest, so \(\( u = 0 \)\). Substituting the values into the first equation:
\(\[
v^2 = 0 + 2 \cdot 2 \cdot 49
\]\)
\(\[
v^2 = 2 \cdot 2 \cdot 49
\]\)
\(\[
v^2 = 4 \cdot 49
\]\)
\(\[
v^2 = 196
\]\)
Taking the square root of both sides to solve for \(\( v \)\):
\(\[
v = \sqrt{196}
\]\)
\(\[
v = 14 \text{ m/s}
\]\)
So, the final velocity of the car after covering 49 meters is 14 m/s.
### Step 3: Calculate Time
Using the second equation and noting that \(\( u = 0 \)\):
\(\[
s = ut + \frac{1}{2}at^2
\]\)
\(\[
49 = 0 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2
\]\)
\(\[
49 = \frac{1}{2} \cdot 2 \cdot t^2
\]\)
\(\[
49 = t^2
\]\)
Taking the square root of both sides to solve for \(\( t \)\):
\(\[
t = \sqrt{49}
\]\)
\(\[
t = 7 \text{ s}
\]\)
So, the time it takes for the car to cover the distance of 49 meters is 7 seconds.
### Summary:
- The final velocity of the car after covering 49 meters is 14 m/s.
- The time it takes to cover this distance is 7 seconds.
