COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 43P

(a)

To determine

Average acceleration between t=6s and t=11s.

(a)

Expert Solution
Check Mark

Answer to Problem 43P

The average acceleration is 2m/s2.

Explanation of Solution

Write the equation to find the average acceleration.

    aav=vfvitfti

Here, aav is the average acceleration, vf is the final velocity, vi is the initial velocity , tf is the final time and ti is the initial time.

Conclusion:

Substitute 14m/s for vf, 4m/s for vi, 11s for tf and 6s for ti in the above equation to find aav.

    aav=14m/s4m/s11s6s=2m/s2

Thus, the average acceleration is 2m/s2.

(b)

To determine

Average velocity between t=6s and t=11s.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The average velocity is 9m/s.

Explanation of Solution

Write the equation to find the average velocity.

    vav=ΔxΔt        (I)

Here, vav is the average velocity, Δx is the total displacement and Δt is the total time.

The area under velocity time graph will give the displacement. Divide the area under the curve between t=6s and t=11s into two segments. They are given below.

A rectangle with sides t=6sto11s and v=0m/sto4m/s. Its area is 20 m.

A right-angled triangle with sides t=6sto11s and v=4m/sto14m/s. Its area is 25 m.

So, the total displacement during t=6sto11s is 45 m.

Conclusion:

Substitute 45m for Δx and 5s for Δt in the equation for vav.

    vav=45m5s=9m/s

Thus, the average velocity is 9m/s.

(c)

To determine

Average velocity between t=0s and t=20s.

(c)

Expert Solution
Check Mark

Answer to Problem 43P

The average velocity is 9.75m/s.

Explanation of Solution

Divide the area into three segments. They are given below.

A rectangle with sides t=0sto20s and v=0m/sto4m/s. Its area is 80 m.

A right-angled triangle with sides t=6sto11s and v=4m/sto14m/s. Its area is 25 m.

A rectangle with sides t=11sto20s and v=4m/sto14m/s. Its area is 90 m.

So, the total displacement during t=0s to t=20s is 195 m.

Conclusion:

Substitute 195m for Δx and 20s for Δt in the equation for vav.

    vav=195m20s=9.75m/s

Thus, the average velocity is 9.75m/s.

(d)

To determine

The distance travelled from t=10s and t=15s.

(d)

Expert Solution
Check Mark

Answer to Problem 43P

The distance travelled is 69 m.

Explanation of Solution

Area under the graph from t=10s to t=15s will give the distance travelled. Area can be easily calculated using the number of unit squares. There are 68 complete squares and two half squares in the considered time interval. So, the total distance travelled will be 69 m.

Thus, the distance travelled is 69 m.

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Chapter 3 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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