Sure, let's solve this problem step by step.
1. Understand the given values:
- Initial velocity ([tex]\( u \)[/tex]): [tex]\( 50 \, \text{m/s} \)[/tex]
- Final velocity ([tex]\( v \)[/tex]): [tex]\( 0 \, \text{m/s} \)[/tex] (since the body comes to rest)
- Deceleration ([tex]\( a \)[/tex]): [tex]\( -3 \, \text{m/s}^2 \)[/tex] (negative because it is a deceleration)
2. Use the kinematic equation:
The appropriate kinematic equation that relates initial velocity, final velocity, acceleration, and distance is:
[tex]\[
v^2 = u^2 + 2as
\][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration (deceleration in this context)
- [tex]\( s \)[/tex] is the distance
3. Rearrange the equation to solve for distance ([tex]\( s \)[/tex]):
[tex]\[
s = \frac{v^2 - u^2}{2a}
\][/tex]
4. Substitute the known values into the equation:
- [tex]\( v = 0 \, \text{m/s} \)[/tex]
- [tex]\( u = 50 \, \text{m/s} \)[/tex]
- [tex]\( a = -3 \, \text{m/s}^2 \)[/tex]
[tex]\[
s = \frac{0^2 - 50^2}{2 \times (-3)}
\][/tex]
5. Simplify the expression:
- Calculate [tex]\( 50^2 \)[/tex]:
[tex]\[
50^2 = 2500
\][/tex]
- Substitute [tex]\( 2500 \)[/tex] into the equation:
[tex]\[
s = \frac{0 - 2500}{2 \times (-3)} = \frac{-2500}{-6}
\][/tex]
6. Calculate the distance:
- Division of the numerator by the denominator:
[tex]\[
s = \frac{2500}{6} \approx 416.67 \, \text{meters}
\][/tex]
So, the body covers a distance of approximately [tex]\( 416.67 \, \text{meters} \)[/tex] from the time it starts to decelerate to the time it comes to rest.
