When I took physics we used rate of change all the time for velocity and acceleration and I could never understand what was going on, now I completely understand..
Problem
1. A rock is thrown off a 550m cliff, what is its velocity when it hits the ground?
I used the formula S(t)=s0+v0t-1/2gt^2 where s0= original position (550m) v0=velocity at t (0 m/s) and g=the acceleration due to gravity (9.8m/s^2), and the formula v(t)=v0-g(t).
s(t)=550-4.9t^2 (the 4.9 came from taking 1/2 of 9.8.) Because the question asks for the velocity when the rock hits the ground, the height is equal to 0. So I set the equation for s(t) equal to 0.
0=550-4.9t^2, solving for t I got t=square root of (550/4.9).
Since the v(t)=v0-g(t), v(t)=-9.8t
To find the velocity I plugged in the square root of (550/4.9) for t and multiplied it by -9.8. The velocity of the rock when it hits the ground is equal to -103.82 m/s. The number is negative because the rock is falling, therefore decreasing in velocity.
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