An object is moving with an initial velocity of \(20 \mathrm{~m} / \mathrm{s}\). If it is decelerating at \(5 \mathrm{~m} / \mathrm{s}^{2}\) how far does it travel before it stops? a) \(10 \mathrm{~m}\) c) \(30 \mathrm{~m}\) b) \(20 \mathrm{~m}\) d) \(40 \mathrm{~m}\)

Short Answer

Expert verified
_(Ignore the effects of gravity and air resistance)_ Answer: The object travels a distance of 40 meters before stopping.

Step by step solution

01

Identify the given information and formula

We are given the following information: - Initial velocity \((u)=20\) m/s - Deceleration rate \((a)=-5\) m/s² (negative because it is slowing down) The formula we will use to solve this problem is: \(v^2 = u^2 + 2as\)
02

Substitute the given values into the formula

We know that the final velocity \((v)=0\) m/s because the object will stop. Substitute all the given values into the formula: \(0^2 = 20^2 + 2(-5) \cdot s\)
03

Simplify and solve for the distance (s)

Carry out the arithmetic operations and isolate 's' to solve for the distance: \(0 = 400 - 10s\) \(10s = 400\) \(s= \frac{400}{10}\) \(s=40\) m
04

State the final answer

The object travels a distance of \(40\) meters before it stops. Therefore, the correct answer is (d) \(40 \mathrm{~m}\).

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