Differentiation. Need help Badly. Thanks?

Question

distance travelled by the ball in metres is given by................ S=t(7-3/4t)
What is the speed of the ball after 16m?
AND Find the acceleration of the ball. please show work. Thanks a mil..............rob

Answer 1

Mlehaji almost got it right.

If the ball is moving up a ramp, at some time it will stop and turn around. Hence the negative acceleration. And two values of time for 16m, going up the ramp and then down,

Answer 2

Expand the right side
S = 7t-3t^2/4

Differentiate wrt t to get velocity:
v = dS/dt = 7 - 3t/2

Differentiate again to get accelaration:
a = d^2S/dt^2 = -3/2 m/s^2

------------------
S = t(7-3/4t)
Now we're finding how much time the ball takes to move 16m.
16 = 7t - 3t^2/4
3t^2/4 -7t +16=0
3t^2 - 28t + 64=0
(3t-16)(t-4)=0

Which gives two values of t (4 and 16/3), both of which give S=16m. This isn't possible, the ball is moving, so it can't be in the same place at both t=4 and t=16/3.

But when you substitute the values in the velocity equation,
v=7-3t/2, 16/3 gives a velocity of -1 which is not possible, so the right value of t is 4s and the velocity of the ball at t=4 (when S=16m) is 1m/s

So speed is 1m/s and acceleration is -3/2m/s^2 (negative since the speed is decreasing)

Answer 3

S=7t -3/4*t^2
dS/dt=7-3/2*t .If S=16m 16 = 7t-3/4*t^2
So 3t^2-28t +64=0 t=((28+-sqrt(16)/6 t=4 and t= 16/3
This is odd if you refer to the horizontal distance,so I suppose that the formula for S applies for the vertical distance
Let's take t=4s
Ds/dt = speed(vertical) = 7-6 =1m/s
d2s/dt2 =-3/2 So the acceleration is constant =-3/2m/s^2.

For me the formula for S is an odd one.
I'm not sure that it makes any sense.
In any case it can't be the horizontal travel of the ball because there can't be two times for the same horizontal distance

Answer 4

1st find the time at 16m so
solve 16 = t(7-3/4 t)

take 1st derivative of S=t(7-3/4t)
S' = (7-3/4 t)-3/4t now plug in the time FYI it not 16m

take another derivative
S'' = -3/4t - 3/4 and plug in the time