Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 2, Problem 63P

(a)

To determine

The average acceleration of given particle.

(a)

Expert Solution
Check Mark

Answer to Problem 63P

The average acceleration in interval AB is 3.3m/s2 , the average acceleration in interval BC is 0.0m/s2 , the average acceleration in interval CE is 7.50m/s2 .

Explanation of Solution

Given:

The variation of velocity of particle with time is plotted.

Formula used:

Write the expression for average acceleration.

  aavg=vfviΔt ........ (1)

Here, aavg is average acceleration of particle, vf is the final velocity, vi is the initial velocity, and Δt is the time interval.

Calculation:

Substitute 15m/s for vf , 5m/s for vi and 3s for Δt in equation (1)

  aavgAB=1553=3.3m/s2 .

Substitute 15m/s for vf , 15m/s for vi and 3s for Δt in equation (1)

  aavgBC=15153=0.0m/s2 .

Substitute 15m/s for vf , 15m/s for vi and 3s for Δt in equation (1)

  aavgCE=15154=7.50m/s2

Conclusion:

Thus, the average acceleration in interval AB is 3.3m/s2 , the average acceleration in interval BC is 0.0m/s2 , the average acceleration in interval CE is 7.50m/s2 .

(b)

To determine

The positionof particle after given time interval.

(b)

Expert Solution
Check Mark

Answer to Problem 63P

The particle is located at distance 75m from staring point after 10s .

Explanation of Solution

Formula used:

The displacement of the particle from the velocity-time graph is given by the area under the curve of the plot for the particular time interval.

The area under the graph for the interval 03s is a trapezium.

Write the expression for the area under the curve for the first interval.

  ΔxAB=12(v0+v3)(Δt1) ........ (1)

Here, ΔxAB is the displacement in first interval, v0 is the velocity at 0s , v3 is the velocity at 3s and Δt1 is the first time interval.

The area under the graph for the interval 36s is rectangle.

Write the expression for the area under the curve for the second interval.

  ΔxBC=v3(Δt2) ........ (2)

Here, ΔxBC is the displacement in second interval, v3 is the velocity at 3s and Δt2 is the second time interval.

The area under the graph for the interval 68s is triangle.

Write the expression for the area under the curve for the third interval.

  ΔxCD=12(v6(Δt3)) ........ (3)

Here, ΔxCD is the displacement in third interval, v6 is the velocity at 6s and Δt3 is the third time interval.

The area under the graph for the interval 810s is triangle.

Write the expression for the area under the curve for the fourth interval.

  ΔxDE=12(v10(Δt4)) ........ (4)

Here, ΔxDE is the displacement in fourth interval, v10 is the velocity at 10s and Δt4 is the third time interval.

Write the expression for displacement.

  Δx=ΔxAB+ΔxBC+ΔxCD+ΔxDE ........ (5)

Here, Δx is displacement in given time interval.

Calculation:

Substitute 5m/s for v0 , 15m/s for v3 and 3s for Δt1 in equation (1).

  ΔxAB=12(5+15)(3)=30m/s .

Substitute 15m/s for v3 and 3s for Δt2 in equation (2).

  ΔxBC=(153)=45m/s

Substitute 15m/s for v6 and 2s for Δt3 in equation (3).

  ΔxCD=12(152)=15m/s

Substitute 15m/s for v10 and 2s for Δt4 in equation (4).

  ΔxDE=12(152)=15m/s .

Substitute 30m/s for ΔxAB , 45m/s for ΔxBC , 15m/s for ΔxCD and 15m/s for ΔxDE in equation (5).

  Δx=30+45+15+(15)=75m

Conclusion:

Thus the particle is located at distance 75m from staring point after 10s .

(c)

To determine

The plot of displacement of particle as a function of time

(c)

Expert Solution
Check Mark

Answer to Problem 63P

The plot of displacement time graph is shown ion figure 1.

Explanation of Solution

The time interval BC is with constant velocity so in displacement time graph interval of BC will be straight line

  Physics for Scientists and Engineers, Chapter 2, Problem 63P

Conclusion:

Thus, the plot of displacement time graph is shown ion figure 1.

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Chapter 2 Solutions

Physics for Scientists and Engineers

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