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secondary 4 | A Maths
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Need help for part 4 only. Thanks
Date Posted: 3 years ago
Views: 206
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We have to be careful that the presence of a timing for which v = 0 may likely mean that there is a change in direction (just as a graph passes the x-axis to change from positive to negative or vice versa).
Here, we find at there is a change in direction at t = 2.16. As such, we cannot just integrate from t = 0 to t = 3 directly since the negative portion of the integral (from t = 2.16 to t = 3) will cancel out, rather than add on to, the positive portion of the integral (from t = 0 to t = 2.16).
Here, we find at there is a change in direction at t = 2.16. As such, we cannot just integrate from t = 0 to t = 3 directly since the negative portion of the integral (from t = 2.16 to t = 3) will cancel out, rather than add on to, the positive portion of the integral (from t = 0 to t = 2.16).
Date Posted:
3 years ago
I sensed that . But couldn’t proceed by myself. Thank u so much Mr Eric.
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