When can instantaneous velocity equal average velocity during a 10-second trip?

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To find when instantaneous velocity equals average velocity during a 10-second trip, the position function is given as s(t) = 3t² - 3t + 5 cm. The instantaneous velocity is calculated as s'(t) = 6t - 3. To determine the average velocity, one must compute the total change in position over the time interval, which is the difference in position at t=10 seconds and t=0 seconds divided by 10. The average velocity is found to be 57 cm/s, and the goal is to find a time t where the instantaneous velocity matches this value. It is essential to avoid posting the same problem in multiple forums to prevent confusion.
sammiyahc0
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The position of a particle moving in a straight line during a 10–second trip is s(t) = 3t2 − 3t + 5 cm.Find a time t at which the instantaneous velocity is equal to the average velocity for the entire trip
 
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Show us some attempt at least!
 
s'(t)=6t-3=6(10)-3=57s instantenous velocity but I don't know how to get average
 
Please don't post the same problem in different forums.

See, you've confused yourself because there may be different replies in your other post.
 
I still need help.
 
sammiyahc0 said:
s'(t)=6t-3=6(10)-3=57s instantenous velocity but I don't know how to get average

Average velocity is defined as change in position divided by time elapsed.
 
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