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[Mechanics 1] Velocity-Time Differentiation

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1. 5 (iv) is what I don't get. I'm probably missing something obvious.

I differentiated ds/dt and got
v = -2t + 20
Then put in 4 for v so
4 = -2t + 20
2t = 16
t = 8
But the answer must be > 10, and the mark scheme uses -4 instead of 4 for v. Why does it use -4?

Thanks
2. (Original post by DarkEnergy)

5 (iv) is what I don't get. I'm probably missing something obvious.

I differentiated ds/dt and got
v = -2t + 20
Then put in 4 for v so
4 = -2t + 20
2t = 16
t = 8
But the answer must be > 10, and the mark scheme uses -4 instead of 4 for v. Why does it use -4?

Thanks
Because the speed is the magnitude of the velocity, and velocity is what you get from differentiation. Therefore .
3. (Original post by RDKGames)
Because the speed is the magnitude of the velocity, and velocity is what you get from differentiation. Therefore .
Ah yes cheers, but how do you know to use -4 or is it just trying both of them and seeing which one gives a result > 10?
4. (Original post by DarkEnergy)
Ah yes cheers, but how do you know to use -4 or is it just trying both of them and seeing which one gives a result > 10?
You don't. Just plug both in and see what you get. Obviously if then you disregard that answer as our domain is strictly as stated by the information we're given.
5. (Original post by RDKGames)
You don't. Just plug both in and see what you get. Obviously if then you disregard that answer as our domain is strictly as stated by the information we're given.
Thanks, thought there was something more complex behind it!
6. (Original post by DarkEnergy)
Thanks, thought there was something more complex behind it!
Nope. Though if you really don't want to try both values, you could plug in t=10 into the differential and see that the value you get is v=0. Therefore for t>10 you will get v<0 just by inspection of the equation, as the minus grows bigger.

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