s(t)=-16t^2+v(subscript0)t+(subscrpt0)
a silver dollar is dropped from the top of a building that is 1362 feet tall. a) determine the position and velocity f'cn for the coin. b) determine the avg velocity on the interval [1,2] c) find the instantanous velocities when t=1 and t=2 d) find the time required for the coin to reach ground level. e) find the velocity of the coin at impact.
-32 (26.1)
Good grief, feet and inches and yards. That is an old textbook with g=-32ft/second^2 so g/2 = 16
anyway
s = (1/2)gt^2 +Vo t + Ho
so
a) s = -16t^2 + 0 + 1362
ds/dt = velocity = -32 t + 0
and of course acceleration =d^2s/dt^2 = -32
b) find out how far it moved in that one second and divide by one second
s(2) = -16*4 +1362
s(1) = -16*1 +1362
difference = -48 ft in one second so average speed = -48 ft/second beteen t = 1 and t = 2
c) v(1) =-32(1) = -32 ft/s
v(2) = -32(2) = -64 ft/s
d) when will s = 0?
0 = -16t^2 + 0 + 1362
16 t^2 = 1362
t^2 = 681
t = 26.1 seconds
e) -16(26.1) =
D) would actually be approximately 9.23 seconds. 1362/16=85.125
To solve this problem, we need to use the given equation for position as a function of time, s(t) = -16t^2 + v₀t + s₀, where s(t) represents the position at time t, v₀ is the initial velocity, and s₀ is the initial position.
a) To determine the position function, we need to substitute the given values into the equation. Here, the initial position s₀ is 1362 feet because the silver dollar was dropped from the top of a 1362-foot building. Since the coin was dropped, the initial velocity v₀ is 0. So, the position equation becomes:
s(t) = -16t^2 + 0t + 1362
= -16t^2 + 1362
b) To find the average velocity on the interval [1, 2], we need to calculate the change in position over the change in time. Average velocity is given by:
Avg. velocity = (s(2) - s(1))/(2 - 1)
Substitute the values into the equation for position:
Avg. velocity = (-(16(2)^2) + 1362 - (-(16(1)^2) + 1362))/(2 - 1)
= (-64 + 1362 - (-16 + 1362))/1
= (1298 - 1346)/1
= -48 ft/s
c) To find the instantaneous velocity at t = 1 and t = 2, we need to find the derivative of the position function with respect to time (t) and evaluate it at t = 1 and t = 2. The derivative of the position function gives the velocity function.
s'(t) = d/dt(-16t^2 + 1362)
= -32t
For t = 1:
Velocity at t = 1, v(1) = s'(1) = -32(1) = -32 ft/s
For t = 2:
Velocity at t = 2, v(2) = s'(2) = -32(2) = -64 ft/s
d) To find the time required for the coin to reach the ground level, we need to determine when the position, s(t), equals zero. In other words, we need to solve the equation -16t^2 + 1362 = 0 for t.
-16t^2 + 1362 = 0
16t^2 = 1362
t^2 = 1362/16
t^2 = 85.125
t = √85.125
t ≈ 9.23 s
Therefore, it takes approximately 9.23 seconds for the coin to reach the ground.
e) Finally, to find the velocity of the coin at impact (when it reaches the ground), we substitute the value of t = 9.23 into the velocity function:
v(9.23) = s'(9.23) = -32(9.23) ≈ -296.96 ft/s
Hence, the velocity of the coin at impact is approximately -296.96 ft/s.