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๐˜ผ ๐™˜๐™–๐™ง ๐™ฉ๐™ง๐™–๐™ซ๐™š๐™ก๐™จ ๐™–๐™ก๐™ค๐™ฃ๐™œ ๐™– ๐™จ๐™ฉ๐™ง๐™–๐™ž๐™œ๐™๐™ฉ ๐™ก๐™ž๐™ฃ๐™š ๐™–๐™ฉ ๐™– ๐™˜๐™ค๐™ฃ๐™จ๐™ฉ๐™–๐™ฃ๐™ฉ ๐™จ๐™ฅ๐™š๐™š๐™™ ๐™ค๐™› 60.0 ๐™ข๐™ž/๐™ ๐™›๐™ค๐™ง ๐™– ๐™™๐™ž๐™จ๐™ฉ๐™–๐™ฃ๐™˜๐™š ๐™™ ๐™–๐™ฃ๐™™ ๐™ฉ๐™๐™š๐™ฃ ๐™–๐™ฃ๐™ค๐™ฉ๐™๐™š๐™ง ๐™™๐™ž๐™จ๐™ฉ๐™–๐™ฃ๐™˜๐™š ๐™™ ๐™ž๐™ฃ ๐™ฉ๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™™๐™ž๐™ง๐™š๐™˜๐™ฉ๐™ž๐™ค๐™ฃ ๐™–๐™ฉ ๐™–๐™ฃ๐™ค๐™ฉ๐™๐™š๐™ง ๐™˜๐™ค๐™ฃ๐™จ๐™ฉ๐™–๐™ฃ๐™ฉ ๐™จ๐™ฅ๐™š๐™š๐™™. ๐™๐™๐™š ๐™–๐™ซ๐™š๐™ง๐™–๐™œ๐™š ๐™ซ๐™š๐™ก๐™ค๐™˜๐™ž๐™ฉ๐™ฎ ๐™›๐™ค๐™ง ๐™ฉ๐™๐™š ๐™š๐™ฃ๐™ฉ๐™ž๐™ง๐™š ๐™ฉ๐™ง๐™ž๐™ฅ ๐™ž๐™จ 30.0 ๐™ข๐™ž/๐™. U๐™จ๐™š ๐™ฉ๐™๐™ž๐™จ ๐™ž๐™ฃ๐™›๐™ค๐™ง๐™ข๐™–๐™ฉ๐™ž๐™ค๐™ฃ ๐™ฉ๐™ค ๐™–๐™ฃ๐™จ๐™ฌ๐™š๐™ง 21-23 21. What is the constant speed with which the car moved during the second distance ๐˜ฅ?

Question

๐˜ผ ๐™˜๐™–๐™ง ๐™ฉ๐™ง๐™–๐™ซ๐™š๐™ก๐™จ ๐™–๐™ก๐™ค๐™ฃ๐™œ ๐™– ๐™จ๐™ฉ๐™ง๐™–๐™ž๐™œ๐™๐™ฉ ๐™ก๐™ž๐™ฃ๐™š ๐™–๐™ฉ ๐™– ๐™˜๐™ค๐™ฃ๐™จ๐™ฉ๐™–๐™ฃ๐™ฉ ๐™จ๐™ฅ๐™š๐™š๐™™ ๐™ค๐™› 60.0 ๐™ข๐™ž/๐™ ๐™›๐™ค๐™ง ๐™– ๐™™๐™ž๐™จ๐™ฉ๐™–๐™ฃ๐™˜๐™š ๐™™ ๐™–๐™ฃ๐™™ ๐™ฉ๐™๐™š๐™ฃ ๐™–๐™ฃ๐™ค๐™ฉ๐™๐™š๐™ง ๐™™๐™ž๐™จ๐™ฉ๐™–๐™ฃ๐™˜๐™š ๐™™ ๐™ž๐™ฃ ๐™ฉ๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™™๐™ž๐™ง๐™š๐™˜๐™ฉ๐™ž๐™ค๐™ฃ ๐™–๐™ฉ ๐™–๐™ฃ๐™ค๐™ฉ๐™๐™š๐™ง ๐™˜๐™ค๐™ฃ๐™จ๐™ฉ๐™–๐™ฃ๐™ฉ ๐™จ๐™ฅ๐™š๐™š๐™™. ๐™๐™๐™š ๐™–๐™ซ๐™š๐™ง๐™–๐™œ๐™š ๐™ซ๐™š๐™ก๐™ค๐™˜๐™ž๐™ฉ๐™ฎ ๐™›๐™ค๐™ง ๐™ฉ๐™๐™š ๐™š๐™ฃ๐™ฉ๐™ž๐™ง๐™š ๐™ฉ๐™ง๐™ž๐™ฅ ๐™ž๐™จ 30.0 ๐™ข๐™ž/๐™. U๐™จ๐™š ๐™ฉ๐™๐™ž๐™จ ๐™ž๐™ฃ๐™›๐™ค๐™ง๐™ข๐™–๐™ฉ๐™ž๐™ค๐™ฃ ๐™ฉ๐™ค ๐™–๐™ฃ๐™จ๐™ฌ๐™š๐™ง 21-23

  1. What is the constant speed with which the car moved during the second distance ๐˜ฅ?
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Solution

The average speed of the entire trip is given by the total distance divided by the total time. The total distance is 2d (d for the first part of the trip and d for the second part). The average speed is given as 30 mi/h.

The speed during the first part of the trip is given as 60 mi/h. Therefore, the time for the first part of the trip is d/60 hours.

Since the average speed of the entire trip is total distance/total time, we can set up the following equation:

30 = 2d / (d/60 + t2)

where t2 is the time for the second part of the trip.

Solving for t2 gives us:

t2 = 2d/30 - d/60

We know that the speed for the second part of the trip is d/t2. Substituting the expression we found for t2 gives us:

Speed2 = d / (2d/30 - d/60)

Solving this equation will give us the speed for the second part of the trip.

This problem has been solved

Similar Questions

๐˜ผ ๐™˜๐™–๐™ง ๐™ฉ๐™ง๐™–๐™ซ๐™š๐™ก๐™จ ๐™–๐™ก๐™ค๐™ฃ๐™œ ๐™– ๐™จ๐™ฉ๐™ง๐™–๐™ž๐™œ๐™๐™ฉ ๐™ก๐™ž๐™ฃ๐™š ๐™–๐™ฉ ๐™– ๐™˜๐™ค๐™ฃ๐™จ๐™ฉ๐™–๐™ฃ๐™ฉ ๐™จ๐™ฅ๐™š๐™š๐™™ ๐™ค๐™› 60.0 ๐™ข๐™ž/๐™ ๐™›๐™ค๐™ง ๐™– ๐™™๐™ž๐™จ๐™ฉ๐™–๐™ฃ๐™˜๐™š ๐™™ ๐™–๐™ฃ๐™™ ๐™ฉ๐™๐™š๐™ฃ ๐™–๐™ฃ๐™ค๐™ฉ๐™๐™š๐™ง ๐™™๐™ž๐™จ๐™ฉ๐™–๐™ฃ๐™˜๐™š ๐™™ ๐™ž๐™ฃ ๐™ฉ๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™™๐™ž๐™ง๐™š๐™˜๐™ฉ๐™ž๐™ค๐™ฃ ๐™–๐™ฉ ๐™–๐™ฃ๐™ค๐™ฉ๐™๐™š๐™ง ๐™˜๐™ค๐™ฃ๐™จ๐™ฉ๐™–๐™ฃ๐™ฉ ๐™จ๐™ฅ๐™š๐™š๐™™. ๐™๐™๐™š ๐™–๐™ซ๐™š๐™ง๐™–๐™œ๐™š ๐™ซ๐™š๐™ก๐™ค๐™˜๐™ž๐™ฉ๐™ฎ ๐™›๐™ค๐™ง ๐™ฉ๐™๐™š ๐™š๐™ฃ๐™ฉ๐™ž๐™ง๐™š ๐™ฉ๐™ง๐™ž๐™ฅ ๐™ž๐™จ 30.0 ๐™ข๐™ž/๐™. U๐™จ๐™š ๐™ฉ๐™๐™ž๐™จ ๐™ž๐™ฃ๐™›๐™ค๐™ง๐™ข๐™–๐™ฉ๐™ž๐™ค๐™ฃ ๐™ฉ๐™ค ๐™–๐™ฃ๐™จ๐™ฌ๐™š๐™ง 21-23 . What is the average speed for the new trip above?

A car starts from rest and accelerates at a rate of ๐‘Ž๐‘1 = 3.0 ๐‘š๐‘  in a straight line until it reaches aspeed of ๐‘ฃ๐‘“ = 25.0 ๐‘š๐‘  . The car then slows at a constant rate of ๐‘Ž๐‘2 = 2.0 ๐‘š๐‘  until it stops. (2 sig-figs)a.) How much time elapses from start to stop?b.) How far does the car travel from start to stop?

A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00 s. How far does the car travel during those 5.00 seconds before it stops?

Car A๐ด starts from a point O๐‘‚ and moves with constant velocity 9m/s9๐‘š/๐‘ . After 2s2๐‘ , another car B๐ต begins its journey from O๐‘‚ and follows car A๐ด. If car B๐ต starts from rest and moves under constant acceleration 4m/s24๐‘š/๐‘ 2, after how much time and at what distance from O๐‘‚ the cars meet?

A car travels at a uniform speed of 60 kmph for 30 minutes and then at a uniform speed of 90 kmph for the next 40 min. Calculate the total distance travelled by car in km.

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