A body moving at [tex]$50 \, m/s$[/tex] decelerates uniformly at [tex]$3 \, m/s^2$[/tex] until it comes to rest. What distance does it cover from the time it starts to decelerate to the time it comes to rest?



Answer :

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.