QuestionA tennis ball hits a wall at $10.0 \mathrm{~m/s}$ and returns at $8.0 \mathrm{~m/s}$ . Find the average acceleration over $0.045 \mathrm{s}$ .

Studdy Solution

STEP 1

Assumptions1. The initial speed of the ball is $10.0 \mathrm{~m/s}$ . . The final speed of the ball is $-8.0 \mathrm{~m/s}$ (negative because it's moving in the opposite direction).
3. The time the ball is in contact with the wall is $0.045 \mathrm{~s}$ .

STEP 2

We need to find the average acceleration of the ball while in contact with the wall. Acceleration is the change in velocity divided by the change in time. The formula for acceleration is $a = \frac{\Delta v}{\Delta t}$

STEP 3

Here, $\Delta v$ is the change in velocity, which is the final velocity minus the initial velocity. So, we can write $\Delta v = v_f - v_i$

STEP 4

Substitute the given values for the initial and final velocities into the equation $\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s}$

STEP 5

Calculate the change in velocity $\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s} = -18.0 \mathrm{~m/s}$

STEP 6

Now, we can substitute the values of $\Delta v$ and $\Delta t$ into the acceleration formula $a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}}$

STEP 7

Calculate the average acceleration $a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}} = -400 \mathrm{~m/s^2}$ The average acceleration of the ball while in contact with the wall is $-400 \mathrm{~m/s^2}$ .

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QuestionA tennis ball hits a wall at $10.0 \mathrm{~m/s}$ and returns at $8.0 \mathrm{~m/s}$ . Find the average acceleration over $0.045 \mathrm{s}$ .

Studdy Solution

STEP 1

Assumptions1. The initial speed of the ball is $10.0 \mathrm{~m/s}$ . . The final speed of the ball is $-8.0 \mathrm{~m/s}$ (negative because it's moving in the opposite direction).
3. The time the ball is in contact with the wall is $0.045 \mathrm{~s}$ .

STEP 2

We need to find the average acceleration of the ball while in contact with the wall. Acceleration is the change in velocity divided by the change in time. The formula for acceleration is $a = \frac{\Delta v}{\Delta t}$

STEP 3

Here, $\Delta v$ is the change in velocity, which is the final velocity minus the initial velocity. So, we can write $\Delta v = v_f - v_i$

STEP 4

Substitute the given values for the initial and final velocities into the equation $\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s}$

STEP 5

Calculate the change in velocity $\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s} = -18.0 \mathrm{~m/s}$

STEP 6

Now, we can substitute the values of $\Delta v$ and $\Delta t$ into the acceleration formula $a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}}$

STEP 7

Calculate the average acceleration $a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}} = -400 \mathrm{~m/s^2}$ The average acceleration of the ball while in contact with the wall is $-400 \mathrm{~m/s^2}$ .

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QuestionA tennis ball hits a wall at 10.0 m/s10.0 \mathrm{~m/s}10.0 m/s and returns at 8.0 m/s8.0 \mathrm{~m/s}8.0 m/s. Find the average acceleration over 0.045s0.045 \mathrm{s}0.045s.

Studdy Solution

STEP 1

Assumptions1. The initial speed of the ball is 10.0 m/s10.0 \mathrm{~m/s}10.0 m/s. . The final speed of the ball is −8.0 m/s-8.0 \mathrm{~m/s}−8.0 m/s (negative because it's moving in the opposite direction).3. The time the ball is in contact with the wall is 0.045 s0.045 \mathrm{~s}0.045 s.

STEP 2

We need to find the average acceleration of the ball while in contact with the wall. Acceleration is the change in velocity divided by the change in time. The formula for acceleration isa=ΔvΔta = \frac{\Delta v}{\Delta t}a=ΔtΔv​

STEP 3

Here, Δv\Delta vΔv is the change in velocity, which is the final velocity minus the initial velocity. So, we can writeΔv=vf−vi\Delta v = v_f - v_iΔv=vf​−vi​

STEP 4

Substitute the given values for the initial and final velocities into the equationΔv=−8.0 m/s−10.0 m/s\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s}Δv=−8.0 m/s−10.0 m/s

STEP 5

Calculate the change in velocityΔv=−8.0 m/s−10.0 m/s=−18.0 m/s\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s} = -18.0 \mathrm{~m/s}Δv=−8.0 m/s−10.0 m/s=−18.0 m/s

STEP 6

Now, we can substitute the values of Δv\Delta vΔv and Δt\Delta tΔt into the acceleration formulaa=−18.0 m/s0.045 sa = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}}a=0.045 s−18.0 m/s​

STEP 7

Calculate the average accelerationa=−18.0 m/s0.045 s=−400 m/s2a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}} = -400 \mathrm{~m/s^2}a=0.045 s−18.0 m/s​=−400 m/s2The average acceleration of the ball while in contact with the wall is −400 m/s2-400 \mathrm{~m/s^2}−400 m/s2.

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QuestionA tennis ball hits a wall at $10.0 \mathrm{~m/s}$ and returns at $8.0 \mathrm{~m/s}$ . Find the average acceleration over $0.045 \mathrm{s}$ .

Assumptions1. The initial speed of the ball is $10.0 \mathrm{~m/s}$ . . The final speed of the ball is $-8.0 \mathrm{~m/s}$ (negative because it's moving in the opposite direction).
3. The time the ball is in contact with the wall is $0.045 \mathrm{~s}$ .

We need to find the average acceleration of the ball while in contact with the wall. Acceleration is the change in velocity divided by the change in time. The formula for acceleration is $a = \frac{\Delta v}{\Delta t}$

Here, $\Delta v$ is the change in velocity, which is the final velocity minus the initial velocity. So, we can write $\Delta v = v_f - v_i$

Substitute the given values for the initial and final velocities into the equation $\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s}$

Calculate the change in velocity $\Delta v = -8.0 \mathrm{~m/s} -10.0 \mathrm{~m/s} = -18.0 \mathrm{~m/s}$

Now, we can substitute the values of $\Delta v$ and $\Delta t$ into the acceleration formula $a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}}$

Calculate the average acceleration $a = \frac{-18.0 \mathrm{~m/s}}{0.045 \mathrm{~s}} = -400 \mathrm{~m/s^2}$ The average acceleration of the ball while in contact with the wall is $-400 \mathrm{~m/s^2}$ .