Differentiation application
Not sure if I post this in the right thread but still hope you guys can help me out with this :-)
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A body moves along a horizontal line according to s=f(t)=t^3 - 9.t^2 +24t where s is the displacement ans t is the time.
1. When is s increasing and when is it decreasing?
2. When is the velocity v increasing, and when it is decreasing?
3. Find the total distance travelled in the first 5 seconds of motion.
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I have no trouble with the 1st one. With the 2nd one I suppose I should use the function v= ds/dt= 3t^2 -18t +24, hence dv/dt= 6t - 18 (please correct me if I'm wrong)
But the 3rd one makes me confused as I don't see where the unit (second) comes from, and if I just apply t=5 to s=f(t), isn't it a little bit too easy?
Thank you guys
P/s: this one is in Singapore's NTU sample paper, given its fame within our region I don't think there is anything wrong with it.
Re: Differentiation application
Quote:
Originally Posted by
kopklan 
A body moves along a horizontal line according to s=f(t)=t^3 - 9.t^2 +24t where s is the displacement ans t is the time.
1. When is s increasing and when is it decreasing?
2. When is the velocity v increasing, and when it is decreasing?
3. Find the total distance travelled in the first 5 seconds of motion.
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My guess is that it should have said something like "meters per second."
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Re: Differentiation application
Quote:
Originally Posted by
kopklan Not sure if I post this in the right thread but still hope you guys can help me out with this :-)
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A body moves along a horizontal line according to s=f(t)=t^3 - 9.t^2 +24t where s is the displacement ans t is the time.
They should have said "t is the time in seconds" in order that 3 make any sense.
Quote:
1. When is s increasing and when is it decreasing?
2. When is the velocity v increasing, and when it is decreasing?
3. Find the total distance travelled in the first 5 seconds of motion.
---
I have no trouble with the 1st one. With the 2nd one I suppose I should use the function v= ds/dt= 3t^2 -18t +24, hence dv/dt= 6t - 18 (please correct me if I'm wrong)
Yes, that is correct (in units, of course, of "distance divided by time").
Quote:
But the 3rd one makes me confused as I don't see where the unit (second) comes from, and if I just apply t=5 to s=f(t), isn't it a little bit too easy?
Yes, too easy and wrong! Notice that v= 3(t^2- 6t+ 8)= 3(t- 2)(t- 4). If t< 2, v is positive. This object moves from s= 0 to s= 8- 9(4)+ 48= 20 a distance of 20 whatever units. From t= 2 to t= 4, v is negative. This object moves from s= 20 to s= 64- 9(16)+ 96= 16, a distance (backwards) of 4 units. From t= 4 to 5, v is positive again. This object moves from s= 16 to s= 125- 9(25)+ 120= 20, a distance of 4 units. The "displacement", the straight line distance from its starting point to its end point is s(5)- s(0)= 20, but the distance move (the distance that would read on its odometer!) is 20+ 4+ 4= 28 of whatever distance units are used.
Quote:
Thank you guys
P/s: this one is in Singapore's NTU sample paper, given its fame within our region I don't think there is anything wrong with it.
