the function is v(t)=50(1-0.9^t)
How do i find the range and how would i solve the equation for t if the velocity is 30 ft per second
How do i find the range and how would i solve the equation for t if the velocity is 30 ft per second
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lim 50(1 - 0.9^t) = 50
t-> ∞
lim 50(1 - 0.9^t) = -∞
t-> -∞
Range is (-∞, 50).
30 = 50(1 - 0.9^t)
0.6 = 1 - 0.9^t
-0.4 = -0.9^t
0.4 = 0.9^t
ln(0.4) = ln(0.9^t)
ln(0.4) = tln(0.9)
t = ln(0.4)/ln(0.9)
t-> ∞
lim 50(1 - 0.9^t) = -∞
t-> -∞
Range is (-∞, 50).
30 = 50(1 - 0.9^t)
0.6 = 1 - 0.9^t
-0.4 = -0.9^t
0.4 = 0.9^t
ln(0.4) = ln(0.9^t)
ln(0.4) = tln(0.9)
t = ln(0.4)/ln(0.9)
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Not clear if particle or car or projectile.
Not specified if 30 ft per second is final target velocity
Short on basic description/info but will try to help.
Standard result: Integral [a^(t)]dt = [a^(t)]/ln a + C
speed = rate of change of distance
dS/dt = v(t) =50 - 50*0.9^t
S = Integral [v(t)]dt = 50 t - (50/ln(0.9))*0.9^t + C
Do we assume starts from rest so S = 0 when t = 0 and so C = 0?
How to solve the equation for t if the v is 30 ft per second
Input as written and command solve. Process is equivalent to:-
50*0.9^t = 50 - v = 20
0.9^t = 20/50 = 0.4
t ln (0.9) = ln(0.4)
t = ln(0.4)/ln (0.9) ~ 8.69671 sec
How to find the range if given
S = 50 t + (474.561)*0.9^t
Not clear since t =infinity yields S = infinity
Perhaps this might mean range object reached when v = 30 and t = 8.69671 sec
If this is the case, (check full description of question), substitute t to get
Range = 624.66 ft
Alternatively this might refer to the range of a function - see other posts
Please include full details in any further posts.
Regards - Ian
Not specified if 30 ft per second is final target velocity
Short on basic description/info but will try to help.
Standard result: Integral [a^(t)]dt = [a^(t)]/ln a + C
speed = rate of change of distance
dS/dt = v(t) =50 - 50*0.9^t
S = Integral [v(t)]dt = 50 t - (50/ln(0.9))*0.9^t + C
Do we assume starts from rest so S = 0 when t = 0 and so C = 0?
How to solve the equation for t if the v is 30 ft per second
Input as written and command solve. Process is equivalent to:-
50*0.9^t = 50 - v = 20
0.9^t = 20/50 = 0.4
t ln (0.9) = ln(0.4)
t = ln(0.4)/ln (0.9) ~ 8.69671 sec
How to find the range if given
S = 50 t + (474.561)*0.9^t
Not clear since t =infinity yields S = infinity
Perhaps this might mean range object reached when v = 30 and t = 8.69671 sec
If this is the case, (check full description of question), substitute t to get
Range = 624.66 ft
Alternatively this might refer to the range of a function - see other posts
Please include full details in any further posts.
Regards - Ian
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Your velocity function v(t) = 50 - 50*(0.9^t) is an exponential function. The DOMAIN is going to restricted to t >= 0, or [0, +infinity) seconds.
Substituting t = 0 into the function: v(0) = 50 - 50*1 = 0ft/s <----This is the lower level restriction on the RANGE since t >= 0 seconds
v(1) = 5 ft/s
v(2) = 9.5 ft/s
v(3) = 13.55 ft/s
v(4) = 17.195 ft/s
v(5) = 20.476 ft/s
...
As you can see the change in the velocity is decreasing and will eventually peak. So, we'll take the limit as t ----> infinity of the velocity function to see where the upper limit lies.
lim t-->infinity [50 - 50*(0.9^t)] = lim t-->infinity [50 - 50*(9/10)^t] = lim t-->infinity 50 = 50 ft/s
Therefore, the RANGE restriction is [0, 50] ft/s
So, v(t) = 30 ft/s
30 = 50(1 - 0.9^t)
0.6 = 1 - 0.9^t
0.9^t = 0.4
ln(0.9^t) = ln(0.4)
tln(0.9) = ln(0.4)
t = ln(0.4)/ln(0.9)
Substituting t = 0 into the function: v(0) = 50 - 50*1 = 0ft/s <----This is the lower level restriction on the RANGE since t >= 0 seconds
v(1) = 5 ft/s
v(2) = 9.5 ft/s
v(3) = 13.55 ft/s
v(4) = 17.195 ft/s
v(5) = 20.476 ft/s
...
As you can see the change in the velocity is decreasing and will eventually peak. So, we'll take the limit as t ----> infinity of the velocity function to see where the upper limit lies.
lim t-->infinity [50 - 50*(0.9^t)] = lim t-->infinity [50 - 50*(9/10)^t] = lim t-->infinity 50 = 50 ft/s
Therefore, the RANGE restriction is [0, 50] ft/s
So, v(t) = 30 ft/s
30 = 50(1 - 0.9^t)
0.6 = 1 - 0.9^t
0.9^t = 0.4
ln(0.9^t) = ln(0.4)
tln(0.9) = ln(0.4)
t = ln(0.4)/ln(0.9)