Average Velocity Formula Calculus

CALCULUS I -NOTES Name __________________________________ Example: The distance of a rock blown upwards from dynamite is given by the equation s(t) = 160t –16t2 a) What is the height of the rock after 2 sec? 3 sec? b) What is the average velocity over the interval 2  t  3? Average velocity is s/t (this is the only velocity that is not the derivative, because it is not at an instant (not instantaneous)). vavg  Δs  256  336   80  80 ft/sec Δt 23 1 Height is position (s), use given equation and plug in given time for t. s(2) = 160(2) – 16(2)2 = 256 ft s(3) = 160(3) – 16(3)2 = 336 ft c) What is the rocks initial velocity? d) What is the velocity at t = 3? Is it going up or down? Velocity is the derivative! s’(t) = 160 – 32t s’(3) = 160 – 32(3) = 64 ft/sec The rock is going up, since the velocity is positive. Velocity is the derivative! Initial velocity is when t = 0. s’(t) = 160 – 32t s’(0) = 160 – 32(0) = 160 ft/sec e d e) What is the velocity at t = 7? Is it going up or down? Velocity is the derivative! s’(t) = 160 – 32t s’(7) = 160 – 32(7) = -64 ft/sec The rock is going down, since the velocity is negative. f) What is the acceleration of the rock at t = 3? Acceleration is the second derivative! s’(t) = 160 – 32t s’’(t) =– 32 ft/sec2 g g) How long does the rock take to reach its maximum height? The maximum height is the point where the object changes direction. It is also where the slope of the tangent line (derivative) = 0!! s'(t) = 160 – 32t 0 = 160 – 32 t 32t = 160 t = 5 sec. Note: Sometimes you will not be able to factor. You will need to use the Quadratic Formula! h) What is the rock’s maximum height? You will need to find time to do this (see part g); you will not always have part g! s(t) = 160t – 16t2 (t = 5 from part g) s(5) = 160(5) – 16(5)2 = 400 ft i) When is the rock at a height of 256 ft? find t, when s = 256! s(t) = 160t – 16t2 256 = 160t – 16t2 16t2 – 160t + 256 = 0 16(t2 – 10t + 16) = 0 16(t – 8)(t – 2) = 0 t–8=0 t–2=0 t=8 t=2 plug s into equation move everything to the same side pull out GCF (if possible) factor set each factor = 0 solve for 2 Does 2 answers make sense? Yes, it was on its way up at 2 sec. and on its way down at 8 sec. Example: A ball is thrown from the top of a building. Its height is defined as s = 60t –16t2 + 200 a) Notice the 200 at the end of the equation; it means it is starting at a height of 200 ft. OR This is asking for the beginning height, so t = 0 s(t) = 60t – 16t2 + 200 s(0) = 60(0) – 16(0)2 + 200 = 200 ft. b) What is the ball’s height after 2 seconds? 4 seconds? Plug t into original equation (s) s(2) = 60(2) – 16(2)2 + 200 = 256 ft s(4) = 60(4) – 16(4)2 + 200 = 184 ft c) What is the ball’s average velocity over the interval 2  t  4? s/t vavg  Δs  256  184  72  -36 ft/sec Δt 24 -2 d) What is the ball’s velocity at 2 seconds? 4 seconds? velocity is derivative s’(t) = 60 – 32t s’(2) = 60 – 32(2) = -4 ft/sec s’(4) = 60 – 32(4) = -68 ft/sec e) What is the ball’s acceleration at 2 seconds? 4 seconds? acceleration is the second derivative s'(t) = 60 – 32t s’’(2) = -32 ft/sec2 s’’(4) = -32 ft/sec2 f) What is the ball’s maximum height? To find height you need time! The maximum height occurs when the derivative = 0! s'(t) = 60 – 32t 0 = 60 – 32t s(1.875) = 60(1.875) – 16(1.875)2 + 200 = 256.25 ft 32t = 60 (plug the time into original equation (s)) t = 1.875 sec g) When does it hit the ground? You need to find time. If the ball is hitting the ground, that means the height (position) is 0. Use the original equation! s = 60t – 16t2 + 200 0 = 60t – 16t2 + 200 0 = -4(4t2 – 15t – 50) Not factorable!  b  b 2  4ac 15  (15)2  4(4)(-50) 15  1025 15  32.02    2a 2(4) 8 8 15  32.02 15  32.02 x  5.88 x  2.13 (can’t have neg. time) 8 8 x h) With what speed does it hit the ground? Speed is the magnitude of velocity (absolute value of derivative) s'(t) = 60 – 32t (you should realize you need time; we know from part f it hits the ground in 5.88 sec) s’(5.88) = 60 – 32(5.88) s’(5.88) = -128.16 speed = |-128.16| = 128.16 ft/sec