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How long does it take for a bicycle traveling at [tex]7.0 \, \text{m/s}[/tex] to come to a stop if the brakes provide an acceleration of [tex]-3.5 \, \text{m/s}^2[/tex]?

A. [tex]3.5 \, \text{s}[/tex]

B. [tex]7.0 \, \text{s}[/tex]

C. [tex]10.5 \, \text{s}[/tex]

D. [tex]2.0 \, \text{s}[/tex]

Sagot :

To solve the problem of determining how long it takes for a bicycle traveling at an initial velocity of [tex]\(7.0 \, \text{m/s}\)[/tex] to come to a stop with a constant acceleration of [tex]\(-3.5 \, \text{m/s}^2\)[/tex], we'll use the kinematic equation which relates velocity, acceleration, and time.

The kinematic equation we will use is:
[tex]\[ v_f = v_i + at \][/tex]

Where:
- [tex]\(v_f\)[/tex] is the final velocity.
- [tex]\(v_i\)[/tex] is the initial velocity.
- [tex]\(a\)[/tex] is the acceleration.
- [tex]\(t\)[/tex] is the time.

Since the bicycle comes to a stop, the final velocity ([tex]\(v_f\)[/tex]) is [tex]\(0 \, \text{m/s}\)[/tex].

Given:
- [tex]\(v_i = 7.0 \, \text{m/s}\)[/tex]
- [tex]\(a = -3.5 \, \text{m/s}^2\)[/tex]
- [tex]\(v_f = 0 \, \text{m/s}\)[/tex]

Plug these values into the kinematic equation:
[tex]\[ 0 = 7.0 + (-3.5) \cdot t \][/tex]

Next, solve for time ([tex]\(t\)[/tex]):
[tex]\[ 0 = 7.0 - 3.5t \][/tex]
[tex]\[ 3.5t = 7.0 \][/tex]
[tex]\[ t = \frac{7.0}{3.5} \][/tex]
[tex]\[ t = 2.0 \, \text{s} \][/tex]

So, the time it takes for the bicycle to come to a stop is:
[tex]\[ \boxed{2.0 \, \text{s}} \][/tex]

Therefore, the correct option is:
D. [tex]\(2.0 \, \text{s}\)[/tex]
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