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2023-02-26

A car traveling 58.5 km/h is 27.4 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.18 s later. (a) What is the car's constant deceleration magnitude before impact? (b) How fast is the car traveling at impact?

Adrienne Clark

Beginner2023-02-27Added 9 answers

First, change the initial speed of the car from kilometers per hour to meters per second

$58.5\frac{{\overline{){\text{km}}}}}{{\overline{){\text{h}}}}}\cdot \frac{1{\overline{){\text{h}}}}}{60{\overline{){\text{min}}}}}\cdot \frac{1{\overline{){\text{min}}}}}{60{\overline{){\text{s}}}}}\cdot \frac{\text{1000 m}}{1{\overline{){\text{km}}}}}=\text{16.25 m/s}$

You can say that the car's velocity will be equal to its speed because it doesn't alter its direction of motion.

So, you know that the car's velocity when it begins to break is equal to 16.25 m/s. Moreover, you know that it took the car 2.18 seconds to cover a distance of 27.4 meters with constant deceleration.

This means that you can write

$d={v}_{0}\cdot t+\frac{1}{2}\cdot a\cdot {t}^{2}}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}$, where

$d$ - the distance to the barrier;

$v}_{0$ - the initial speed of the car;

$t$ - the stopping time;

$a$ - the deceleration - you can expect it to be negative;

If the direction of motion is assumed to be positive, then the acceleration vector for a car that is slowing down will be orientated in the opposite direction of its motion, therefore you may anticipate the acceleration to be negative.

Rearrange this equation to solve for $a$

$d={v}_{0}\cdot t=\frac{1}{2}\cdot a\cdot {t}^{2}$

$a=\frac{2\cdot (d-{v}_{0}\cdot t)}{{t}^{2}}$

$a=\frac{2\cdot (27.4\text{m}-16.25\frac{\text{m}}{{\overline{){\text{s}}}}}\cdot 2.18{\overline{){\text{s}}}})}{{2.18}^{2}\text{s}{\text{}}^{2}}={-\text{3.38 m/s}{\text{}}^{2}}$

To determine the car's speed upon impact, use the equation

${v}_{\text{impact}}={v}_{0}+a\cdot t}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}$, where

$v}_{\text{impact}$ - the speed upon impact;

Plug in your values to get

$v}_{\text{impact}}=16.25\frac{\text{m}}{\text{s}}+(-3.38)\frac{\text{m}}{{\text{s}}^{{\overline{){2}}}}}\cdot 2.18{\overline{){\text{s}}}}={\text{8.88 m/s}$

$58.5\frac{{\overline{){\text{km}}}}}{{\overline{){\text{h}}}}}\cdot \frac{1{\overline{){\text{h}}}}}{60{\overline{){\text{min}}}}}\cdot \frac{1{\overline{){\text{min}}}}}{60{\overline{){\text{s}}}}}\cdot \frac{\text{1000 m}}{1{\overline{){\text{km}}}}}=\text{16.25 m/s}$

You can say that the car's velocity will be equal to its speed because it doesn't alter its direction of motion.

So, you know that the car's velocity when it begins to break is equal to 16.25 m/s. Moreover, you know that it took the car 2.18 seconds to cover a distance of 27.4 meters with constant deceleration.

This means that you can write

$d={v}_{0}\cdot t+\frac{1}{2}\cdot a\cdot {t}^{2}}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}$, where

$d$ - the distance to the barrier;

$v}_{0$ - the initial speed of the car;

$t$ - the stopping time;

$a$ - the deceleration - you can expect it to be negative;

If the direction of motion is assumed to be positive, then the acceleration vector for a car that is slowing down will be orientated in the opposite direction of its motion, therefore you may anticipate the acceleration to be negative.

Rearrange this equation to solve for $a$

$d={v}_{0}\cdot t=\frac{1}{2}\cdot a\cdot {t}^{2}$

$a=\frac{2\cdot (d-{v}_{0}\cdot t)}{{t}^{2}}$

$a=\frac{2\cdot (27.4\text{m}-16.25\frac{\text{m}}{{\overline{){\text{s}}}}}\cdot 2.18{\overline{){\text{s}}}})}{{2.18}^{2}\text{s}{\text{}}^{2}}={-\text{3.38 m/s}{\text{}}^{2}}$

To determine the car's speed upon impact, use the equation

${v}_{\text{impact}}={v}_{0}+a\cdot t}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}$, where

$v}_{\text{impact}$ - the speed upon impact;

Plug in your values to get

$v}_{\text{impact}}=16.25\frac{\text{m}}{\text{s}}+(-3.38)\frac{\text{m}}{{\text{s}}^{{\overline{){2}}}}}\cdot 2.18{\overline{){\text{s}}}}={\text{8.88 m/s}$