Question

The position of a softball tossed vertically upward is described by the equation y=c1(t)−c2(t^2), wherey is in meters,tin seconds,c1=6.71 m/s, and c2= 2.39 m/s^2. How would you find the acceleration at a specific t value?

Answer 1

\[\Large y=c_1t-c_2t^2\] `Position` is usually denoted by s as a function of t:\[\Large s(t)=c_1t-c_2t^2\]

The instantaneous rate of change of postion gives us velocity,\[\Large s'(t)=v(t)\]The second derivative of position,
or the first derivative of velocity,
gives us acceleration.\[\Large s''(t)=a(t)\]
Then to find the acceleration at a specific \(\Large t\) value, let's say at \(\Large b\), we would plug that value into the acceleration function.
\[\Large s''(b)=a(b)\]

Answer 2

So it looks like you'll want to find the second derivative of your function :o
Understand how to do that?

Answer 3

I did that and got 4.78 but apparently that is wrong

Answer 4

\[\Large s(t)=c_1t-c_2t^2\]\[\Large s'(t)=c_1-2c_2t\]\[\Large s''(t)=-2c_2\]
Hmm it looks like you've got the right idea :)
Did you remember the negative sign?

Answer 5

nope! lol

Answer 6

Always the little things :/

Answer 7

Hehe yah ^^

Answer 8

thank you :)

Answer 9

lol

Answer 10

hey

Answer 11

hi

Answer 12

hi

Answer 13

watcha doubt
its tough rite

Answer 14

do u mind givin me a medal plzz

Answer 15

watcha doubt
its tough rite

Answer 16

hi