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secondary 4 | A Maths
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Date Posted: 4 years ago
Views: 213
2 am if no one does this yet
Distance s = 3t + 4sin(2t)
For part (i) sub in value of t = 1.1 into s
Velocity = differentiate of distance
s’ = 3 + 8cos(2t)
For part (ii) sub in value of t = pi/6 into s’
Acceleration = differentiate of velocity
s” = -16sin(2t)
Question asked when it stops which means velocity = 0
Sub in s’ = 0 to find the time (t) when this happens.
After getting t, sub it into s” to get your acceleration or deceleration.
Sorry I’m out now and don’t have access to a proper calculator.
For part (i) sub in value of t = 1.1 into s
Velocity = differentiate of distance
s’ = 3 + 8cos(2t)
For part (ii) sub in value of t = pi/6 into s’
Acceleration = differentiate of velocity
s” = -16sin(2t)
Question asked when it stops which means velocity = 0
Sub in s’ = 0 to find the time (t) when this happens.
After getting t, sub it into s” to get your acceleration or deceleration.
Sorry I’m out now and don’t have access to a proper calculator.
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Date Posted:
4 years ago
It’s in cm not m
Oops
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