QuestionA bicyclist at $14 \mathrm{~m/s}$ accelerates at $2 \mathrm{~m/s^2}$ down a 20 m hill. Find her speed at the bottom.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the bicyclist is $14 \mathrm{~m/s}$ . The acceleration of the bicyclist is \mathrm{~m/s^}
3. The distance of the hill is $20 \mathrm{~m}$
4. We need to find the final velocity of the bicyclist at the bottom of the hill

STEP 2

We can use the equation of motion to solve this problem. The equation is $v_f^2 = v_i^2 +2a d$ where- $v_f$ is the final velocity- $v_i$ is the initial velocity- $a$ is the acceleration- $d$ is the distance

STEP 3

Now, plug in the given values for the initial velocity, acceleration, and distance into the equation.
$v_f^2 =14^2 +2 \cdot2 \cdot20$

STEP 4

Calculate the right side of the equation.
$v_f^2 =196 +80$

STEP 5

Add the two values together.
$v_f^2 =276$

STEP 6

To find the final velocity, we need to take the square root of both sides of the equation.
$v_f = \sqrt{276}$

STEP 7

Calculate the final velocity.
$v_f = \sqrt{276} \approx16.6 \mathrm{~m/s}$ The bicyclist is going approximately $16.6 \mathrm{~m/s}$ at the bottom of the hill.

Was this helpful?

QuestionA bicyclist at $14 \mathrm{~m/s}$ accelerates at $2 \mathrm{~m/s^2}$ down a 20 m hill. Find her speed at the bottom.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the bicyclist is $14 \mathrm{~m/s}$ . The acceleration of the bicyclist is \mathrm{~m/s^}
3. The distance of the hill is $20 \mathrm{~m}$
4. We need to find the final velocity of the bicyclist at the bottom of the hill

STEP 2

We can use the equation of motion to solve this problem. The equation is $v_f^2 = v_i^2 +2a d$ where- $v_f$ is the final velocity- $v_i$ is the initial velocity- $a$ is the acceleration- $d$ is the distance

STEP 3

Now, plug in the given values for the initial velocity, acceleration, and distance into the equation.
$v_f^2 =14^2 +2 \cdot2 \cdot20$

STEP 4

Calculate the right side of the equation.
$v_f^2 =196 +80$

STEP 5

Add the two values together.
$v_f^2 =276$

STEP 6

To find the final velocity, we need to take the square root of both sides of the equation.
$v_f = \sqrt{276}$

STEP 7

Calculate the final velocity.
$v_f = \sqrt{276} \approx16.6 \mathrm{~m/s}$ The bicyclist is going approximately $16.6 \mathrm{~m/s}$ at the bottom of the hill.

Get Studdy

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.

QuestionA bicyclist at 14 m/s14 \mathrm{~m/s}14 m/s accelerates at 2 m/s22 \mathrm{~m/s^2}2 m/s2 down a 20 m hill. Find her speed at the bottom.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the bicyclist is 14 m/s14 \mathrm{~m/s}14 m/s . The acceleration of the bicyclist is \mathrm{~m/s^}3. The distance of the hill is 20 m20 \mathrm{~m}20 m4. We need to find the final velocity of the bicyclist at the bottom of the hill

STEP 2

We can use the equation of motion to solve this problem. The equation isvf2=vi2+2adv_f^2 = v_i^2 +2a dvf2​=vi2​+2adwhere- vfv_fvf​ is the final velocity- viv_ivi​ is the initial velocity- aaa is the acceleration- ddd is the distance

STEP 3

Now, plug in the given values for the initial velocity, acceleration, and distance into the equation.vf2=142+2⋅2⋅20v_f^2 =14^2 +2 \cdot2 \cdot20vf2​=142+2⋅2⋅20

STEP 4

Calculate the right side of the equation.vf2=196+80v_f^2 =196 +80vf2​=196+80

STEP 5

Add the two values together.vf2=276v_f^2 =276vf2​=276

STEP 6

To find the final velocity, we need to take the square root of both sides of the equation.vf=276v_f = \sqrt{276}vf​=276​

STEP 7

Calculate the final velocity.vf=276≈16.6 m/sv_f = \sqrt{276} \approx16.6 \mathrm{~m/s}vf​=276​≈16.6 m/sThe bicyclist is going approximately 16.6 m/s16.6 \mathrm{~m/s}16.6 m/s at the bottom of the hill.

Get Studdy

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.

QuestionA bicyclist at $14 \mathrm{~m/s}$ accelerates at $2 \mathrm{~m/s^2}$ down a 20 m hill. Find her speed at the bottom.

Assumptions1. The initial velocity of the bicyclist is $14 \mathrm{~m/s}$ . The acceleration of the bicyclist is \mathrm{~m/s^}
3. The distance of the hill is $20 \mathrm{~m}$
4. We need to find the final velocity of the bicyclist at the bottom of the hill

We can use the equation of motion to solve this problem. The equation is $v_f^2 = v_i^2 +2a d$ where- $v_f$ is the final velocity- $v_i$ is the initial velocity- $a$ is the acceleration- $d$ is the distance

Now, plug in the given values for the initial velocity, acceleration, and distance into the equation.
$v_f^2 =14^2 +2 \cdot2 \cdot20$

Calculate the right side of the equation.
$v_f^2 =196 +80$

Add the two values together.
$v_f^2 =276$

To find the final velocity, we need to take the square root of both sides of the equation.
$v_f = \sqrt{276}$

Calculate the final velocity.
$v_f = \sqrt{276} \approx16.6 \mathrm{~m/s}$ The bicyclist is going approximately $16.6 \mathrm{~m/s}$ at the bottom of the hill.