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Q105P

Expert-verifiedFound in: Page 39

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: A motorcycle is moving at 30 m/s** **when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.0 s** **interval immediately after braking begins, the speed decreases to 15 m/s****. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops?**

The distance traveled by the motorcycle during deceleration is 90 m.

Initial speed is, ${v}_{0=30\text{m/s}}$

Final speed after 3.0 s is, $v=15\text{m/s}$

** **

**Kinematic equations describe the motion of an object with constant acceleration. Using these equations the deceleration of the motorcycle on applying breaks can be computed. This deceleration can be used to find the distance traveled by the motorcycle. **

The expression for the kinematic equations of motion are given as follows:

${\text{v=v}}_{0}\text{+at}\phantom{\rule{0ex}{0ex}}$ … (i)

${\text{v}}^{2}{{\text{= v}}_{0}}^{2}\text{+2ax}$

… (ii)

Here,${\text{v}}_{0}$ is the initial velocity, $t$ is the time, $a$is the acceleration and $x$ is the displacement.

Using equation (i), the acceleration can be calculated as,

$\text{a=}\frac{{\text{v-v}}_{0}}{t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{=}\frac{15\text{m/s-30m//s}}{3.0\text{s}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{= -5m/s}}^{2}$

Since acceleration is negative, it is called deceleration

Using the equation (ii) the distance traveled can be calculated as,

$x=\frac{{v}^{2}-{{v}_{0}}^{2}}{2a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(0\text{m/s}\right)}^{2}-{\left(30\text{m/s}\right)}^{2}}{2\times \left(-5{\text{m/s}}^{2}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=90\text{m}$

Thus the distance travelled during deceleration is $90\text{m}$ .

** **

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