Velocity of a Raindrop: Solving $\frac{dv}{dt} = 9 - 09.t$

[cscott]

A raindrop has an initial downward speed of 10 m/s and its downward acceleration is given by
a = \left \{ \begin{array}{cc} 9 - 0.9t, &amp;0 \le t \le 10<br /> \\0, &amp;t \ge 10\end{array}\right.

What is the velocity after 1s? (55 m/s)

I did

\frac{dv}{dt} = 9 - 09.t \Leftrightarrow v = 9t - 0.45t^2 + C

when
t = 0, v = -10 = 9(0) - 0.45(0)^2 + C \Leftrightarrow C = -10

But with this I get -1.45.

 

[greytomato]

i think you got a mistype there...
if you calculate the speed after 10s you should get 55 m/s.
also notice the sign of C.

 

[cscott]

Says 1 s in the book, so I guess it is a typo.