QuestionA car traveling at $24 \mathrm{~ms}^{-1}$ stops after $19.2 \mathrm{~m}$ . Find the average acceleration during braking.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the car is $24 \mathrm{~ms}^{-1}$ . The final velocity of the car is $0 \mathrm{~ms}^{-1}$ (since the car stops)
3. The distance travelled by the car while braking is $19. \mathrm{~m}$
4. We are asked to find the average acceleration, which is the change in velocity divided by the time taken. However, we don't have the time. But we can use the kinematic equation to solve for acceleration.

STEP 2

We can use the second equation of motion to find the acceleration, which is $v^2 = u^2 +2as$ where- $v$ is the final velocity- $u$ is the initial velocity- $a$ is the acceleration- $s$ is the distance travelled

STEP 3

Substitute the given values into the equation.
$0 = (24)^2 +2a(19.2)$

STEP 4

Rearrange the equation to solve for acceleration.
$2a(19.2) = -(24)^2$

STEP 5

olve for acceleration.
$a = \frac{-(24)^2}{2(19.2)}$

STEP 6

Calculate the acceleration.
$a = \frac{-(24)^2}{2(19.2)} = -15 \mathrm{~ms}^{-2}$ The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is expected as the car is braking.
The average acceleration of the car during braking is $-15 \mathrm{~ms}^{-2}$ .

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QuestionA car traveling at $24 \mathrm{~ms}^{-1}$ stops after $19.2 \mathrm{~m}$ . Find the average acceleration during braking.

Studdy Solution

STEP 1

STEP 2

We can use the second equation of motion to find the acceleration, which is $v^2 = u^2 +2as$ where- $v$ is the final velocity- $u$ is the initial velocity- $a$ is the acceleration- $s$ is the distance travelled

STEP 3

Substitute the given values into the equation.
$0 = (24)^2 +2a(19.2)$

STEP 4

Rearrange the equation to solve for acceleration.
$2a(19.2) = -(24)^2$

STEP 5

olve for acceleration.
$a = \frac{-(24)^2}{2(19.2)}$

STEP 6

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QuestionA car traveling at 24 ms−124 \mathrm{~ms}^{-1}24 ms−1 stops after 19.2 m19.2 \mathrm{~m}19.2 m. Find the average acceleration during braking.

Studdy Solution

STEP 1

STEP 2

We can use the second equation of motion to find the acceleration, which isv2=u2+2asv^2 = u^2 +2asv2=u2+2aswhere- vvv is the final velocity- uuu is the initial velocity- aaa is the acceleration- sss is the distance travelled

STEP 3

Substitute the given values into the equation.0=(24)2+2a(19.2)0 = (24)^2 +2a(19.2)0=(24)2+2a(19.2)

STEP 4

Rearrange the equation to solve for acceleration.2a(19.2)=−(24)22a(19.2) = -(24)^22a(19.2)=−(24)2

STEP 5

olve for acceleration.a=−(24)22(19.2)a = \frac{-(24)^2}{2(19.2)}a=2(19.2)−(24)2​

STEP 6

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QuestionA car traveling at $24 \mathrm{~ms}^{-1}$ stops after $19.2 \mathrm{~m}$ . Find the average acceleration during braking.

We can use the second equation of motion to find the acceleration, which is $v^2 = u^2 +2as$ where- $v$ is the final velocity- $u$ is the initial velocity- $a$ is the acceleration- $s$ is the distance travelled

Substitute the given values into the equation.
$0 = (24)^2 +2a(19.2)$

Rearrange the equation to solve for acceleration.
$2a(19.2) = -(24)^2$

olve for acceleration.
$a = \frac{-(24)^2}{2(19.2)}$