Math / Algebra

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A car is traveling at $112 \mathrm{~km} / \mathrm{h}$ when the driver sees an accident 80 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup? (Round your answer to two decimal places.)

Studdy Solution

STEP 1

What is this asking? How hard does a driver need to brake to avoid hitting something 80 meters ahead while driving at 112 km/h? Watch out! The units are different!
We've got kilometers per hour and meters.
Mixing units is a recipe for disaster!

STEP 2

1. Convert Units
2. Set Up the Equation
3. Solve for Deceleration

STEP 3

Alright, let's get those units matching!
We need to convert the speed from km/h to m/s.
Why? Because the distance is given in meters.
Consistency is key!

STEP 4

We know that 1 km is equal to 1000 m, and 1 hour is equal to 3600 seconds.
So, to convert 112 km/h to m/s, we multiply by 1000 m/km and divide by 3600 s/h.
It's like magic!

STEP 5

$112 \frac{\text{km}}{\text{h}} \cdot \frac{1000 \text{ m}}{1 \text{ km}} \cdot \frac{1 \text{ h}}{3600 \text{ s}} = \frac{112 \cdot 1000}{3600} \frac{\text{m}}{\text{s}} = \frac{1120}{36} \frac{\text{m}}{\text{s}} \approx 31.11 \frac{\text{m}}{\text{s}}$

STEP 6

Boom! Our initial velocity, $v_0$ , is approximately $31.11$ m/s.
Much better!

STEP 7

We're dealing with constant deceleration, which means constant acceleration in the opposite direction of motion.
We can use this handy-dandy equation of motion:

STEP 8

$v^2 = v_0^2 + 2 \cdot a \cdot d$

STEP 9

Where: $v$ is the final velocity (which is 0 m/s since we want to stop), $v_0$ is the initial velocity (that's our $31.11$ m/s), $a$ is the acceleration (which is what we're looking for, our deceleration will be $-a$ ), and $d$ is the distance (80 m).

STEP 10

Let's plug in what we know!

STEP 11

$0^2 = (31.11)^2 + 2 \cdot a \cdot 80$

STEP 12

Now, we solve for $a$ :

STEP 13

$0 = 967.8321 + 160a$ $-967.8321 = 160a$ $a = \frac{-967.8321}{160}$ $a \approx -6.05 \frac{\text{m}}{\text{s}^2}$

STEP 14

Since we're looking for deceleration, we take the positive value.
So, the minimum constant deceleration required is approximately $6.05$ m/s².

STEP 15

The minimum constant deceleration required to stop the car in time is approximately $6.05$ m/s².

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Studdy Solution

STEP 1

What is this asking? How hard does a driver need to brake to avoid hitting something 80 meters ahead while driving at 112 km/h? Watch out! The units are different!
We've got kilometers per hour and meters.
Mixing units is a recipe for disaster!

STEP 2

1. Convert Units
2. Set Up the Equation
3. Solve for Deceleration

STEP 3

Alright, let's get those units matching!
We need to convert the speed from km/h to m/s.
Why? Because the distance is given in meters.
Consistency is key!

STEP 4

We know that 1 km is equal to 1000 m, and 1 hour is equal to 3600 seconds.
So, to convert 112 km/h to m/s, we multiply by 1000 m/km and divide by 3600 s/h.
It's like magic!

STEP 5

$112 \frac{\text{km}}{\text{h}} \cdot \frac{1000 \text{ m}}{1 \text{ km}} \cdot \frac{1 \text{ h}}{3600 \text{ s}} = \frac{112 \cdot 1000}{3600} \frac{\text{m}}{\text{s}} = \frac{1120}{36} \frac{\text{m}}{\text{s}} \approx 31.11 \frac{\text{m}}{\text{s}}$

STEP 6

Boom! Our initial velocity, $v_0$ , is approximately $31.11$ m/s.
Much better!

STEP 7

We're dealing with constant deceleration, which means constant acceleration in the opposite direction of motion.
We can use this handy-dandy equation of motion:

STEP 8

$v^2 = v_0^2 + 2 \cdot a \cdot d$

STEP 9

Where: $v$ is the final velocity (which is 0 m/s since we want to stop), $v_0$ is the initial velocity (that's our $31.11$ m/s), $a$ is the acceleration (which is what we're looking for, our deceleration will be $-a$ ), and $d$ is the distance (80 m).

STEP 10

Let's plug in what we know!

STEP 11

$0^2 = (31.11)^2 + 2 \cdot a \cdot 80$

STEP 12

Now, we solve for $a$ :

STEP 13

$0 = 967.8321 + 160a$ $-967.8321 = 160a$ $a = \frac{-967.8321}{160}$ $a \approx -6.05 \frac{\text{m}}{\text{s}^2}$

STEP 14

Since we're looking for deceleration, we take the positive value.
So, the minimum constant deceleration required is approximately $6.05$ m/s².

STEP 15

The minimum constant deceleration required to stop the car in time is approximately $6.05$ m/s².

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Studdy Solution

STEP 1

What is this asking? How hard does a driver need to brake to avoid hitting something 80 meters ahead while driving at 112 km/h? Watch out! The units are different!We've got kilometers per hour and meters.Mixing units is a recipe for disaster!

STEP 2

1. Convert Units2. Set Up the Equation3. Solve for Deceleration

STEP 3

Alright, let's get those units matching!We need to convert the speed from km/h to m/s.Why? Because the distance is given in meters.Consistency is key!

STEP 4

We know that 1 km is equal to 1000 m, and 1 hour is equal to 3600 seconds.So, to convert 112 km/h to m/s, we **multiply** by 1000 m/km and **divide** by 3600 s/h.It's like magic!

STEP 5

STEP 6

Boom! Our **initial velocity**, v0v_0v0​, is approximately 31.1131.1131.11 m/s.Much better!

STEP 7

We're dealing with constant deceleration, which means constant acceleration in the opposite direction of motion.We can use this handy-dandy equation of motion:

STEP 8

v2=v02+2⋅a⋅dv^2 = v_0^2 + 2 \cdot a \cdot dv2=v02​+2⋅a⋅d

STEP 9

Where: vvv is the **final velocity** (which is 0 m/s since we want to stop), v0v_0v0​ is the **initial velocity** (that's our 31.1131.1131.11 m/s), aaa is the **acceleration** (which is what we're looking for, our deceleration will be −a-a−a), and ddd is the **distance** (80 m).

STEP 10

Let's plug in what we know!

STEP 11

02=(31.11)2+2⋅a⋅800^2 = (31.11)^2 + 2 \cdot a \cdot 8002=(31.11)2+2⋅a⋅80

STEP 12

Now, we **solve for** aaa:

STEP 13

0=967.8321+160a0 = 967.8321 + 160a0=967.8321+160a −967.8321=160a-967.8321 = 160a−967.8321=160a a=−967.8321160a = \frac{-967.8321}{160}a=160−967.8321​a≈−6.05ms2a \approx -6.05 \frac{\text{m}}{\text{s}^2}a≈−6.05s2m​

STEP 14

Since we're looking for deceleration, we take the positive value.So, the minimum constant deceleration required is approximately 6.056.056.05 m/s².

STEP 15

The minimum constant deceleration required to stop the car in time is approximately **6.056.056.05 m/s²**.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? How hard does a driver need to brake to avoid hitting something 80 meters ahead while driving at 112 km/h? Watch out! The units are different!
We've got kilometers per hour and meters.
Mixing units is a recipe for disaster!

1. Convert Units
2. Set Up the Equation
3. Solve for Deceleration

Alright, let's get those units matching!
We need to convert the speed from km/h to m/s.
Why? Because the distance is given in meters.
Consistency is key!

We know that 1 km is equal to 1000 m, and 1 hour is equal to 3600 seconds.
So, to convert 112 km/h to m/s, we multiply by 1000 m/km and divide by 3600 s/h.
It's like magic!

Boom! Our initial velocity, $v_0$ , is approximately $31.11$ m/s.
Much better!

We're dealing with constant deceleration, which means constant acceleration in the opposite direction of motion.
We can use this handy-dandy equation of motion:

$v^2 = v_0^2 + 2 \cdot a \cdot d$

Where: $v$ is the final velocity (which is 0 m/s since we want to stop), $v_0$ is the initial velocity (that's our $31.11$ m/s), $a$ is the acceleration (which is what we're looking for, our deceleration will be $-a$ ), and $d$ is the distance (80 m).

$0^2 = (31.11)^2 + 2 \cdot a \cdot 80$

Now, we solve for $a$ :

$0 = 967.8321 + 160a$ $-967.8321 = 160a$ $a = \frac{-967.8321}{160}$ $a \approx -6.05 \frac{\text{m}}{\text{s}^2}$

Since we're looking for deceleration, we take the positive value.
So, the minimum constant deceleration required is approximately $6.05$ m/s².

The minimum constant deceleration required to stop the car in time is approximately $6.05$ m/s².