Answer :
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 [tex]\( u \)[/tex] is 0 m/s.
- The acceleration [tex]\( a \)[/tex] is 2 m/s².
- The distance [tex]\( s \)[/tex] covered is 49 meters.
2. Equations Used:
- To find the final velocity ([tex]\( v \)[/tex]), we use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
- To find the time ([tex]\( t \)[/tex]), we use the kinematic equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
### Step 2: Calculate Final Velocity
The car starts from rest, so [tex]\( u = 0 \)[/tex]. Substituting the values into the first equation:
[tex]\[ v^2 = 0 + 2 \cdot 2 \cdot 49 \][/tex]
[tex]\[ v^2 = 2 \cdot 2 \cdot 49 \][/tex]
[tex]\[ v^2 = 4 \cdot 49 \][/tex]
[tex]\[ v^2 = 196 \][/tex]
Taking the square root of both sides to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{196} \][/tex]
[tex]\[ v = 14 \text{ m/s} \][/tex]
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 [tex]\( u = 0 \)[/tex]:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
[tex]\[ 49 = 0 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2 \][/tex]
[tex]\[ 49 = \frac{1}{2} \cdot 2 \cdot t^2 \][/tex]
[tex]\[ 49 = t^2 \][/tex]
Taking the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt{49} \][/tex]
[tex]\[ t = 7 \text{ s} \][/tex]
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.
### Step 1: Understand Initial Conditions and Equations
1. Initial Conditions:
- The car starts from rest, so its initial velocity [tex]\( u \)[/tex] is 0 m/s.
- The acceleration [tex]\( a \)[/tex] is 2 m/s².
- The distance [tex]\( s \)[/tex] covered is 49 meters.
2. Equations Used:
- To find the final velocity ([tex]\( v \)[/tex]), we use the kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
- To find the time ([tex]\( t \)[/tex]), we use the kinematic equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
### Step 2: Calculate Final Velocity
The car starts from rest, so [tex]\( u = 0 \)[/tex]. Substituting the values into the first equation:
[tex]\[ v^2 = 0 + 2 \cdot 2 \cdot 49 \][/tex]
[tex]\[ v^2 = 2 \cdot 2 \cdot 49 \][/tex]
[tex]\[ v^2 = 4 \cdot 49 \][/tex]
[tex]\[ v^2 = 196 \][/tex]
Taking the square root of both sides to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{196} \][/tex]
[tex]\[ v = 14 \text{ m/s} \][/tex]
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 [tex]\( u = 0 \)[/tex]:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
[tex]\[ 49 = 0 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2 \][/tex]
[tex]\[ 49 = \frac{1}{2} \cdot 2 \cdot t^2 \][/tex]
[tex]\[ 49 = t^2 \][/tex]
Taking the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt{49} \][/tex]
[tex]\[ t = 7 \text{ s} \][/tex]
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.
