The motion of an avalanche is described by s1t2 3t2, where s is the distance, in metres, travelled by the leading edge of the snow at t seconds.
a. Find the distance travelled from 0s to 5s.
b. Find the rate at which the avalanche is moving from 0s to10s.
c. Find the rate at which the avalanche is moving at 10s.
d. How long, to the nearest second, does the leading edge of the snow take to move 600 m?
when t = 0, s = 0
when t = 5 , s = 3(25) = 75
b) when t = 10 , s = 300
rate = (300-0)/(10-0) = 30 m/s
c) v(t) = 6t
when t = 10, v(10) = 60 m/s
d) 3t^2 = 600
t^2 = 200
t = √200 = appr 14.1 seconds
a) s(0) =0
s(5) = 3t^2 = 3(5)^2 = 75m
b) s(10) = 3(10)^2 = 300
s(10) - s(0) / 10-0 = 30m/s
c)s(10) = 300 s(10+h) = 3(10+h)^2 = 300+60h+3h^2
s(10 +h) - s(10) / h
= 60 + 3h ; where limit h->0
=60m/s
d)600=3t^2
200 = t^2
t = 14.1 s
the motion of the avaknce is described by s(t)=3t^2
The next three terms
a. To find the distance travelled from 0s to 5s, we substitute t = 5s into the equation:
s(5) = 3(5)^2
s(5) = 3(25)
s(5) = 75m
So, the distance travelled from 0s to 5s is 75m.
b. To find the rate at which the avalanche is moving from 0s to 10s, we use the average rate of change formula:
Rate = (s(10) - s(0))/(10 - 0)
Substituting t = 10s into the equation:
s(10) = 3(10)^2
s(10) = 3(100)
s(10) = 300m
Now we can calculate the rate:
Rate = (300 - 0)/(10 - 0)
Rate = 300/10
Rate = 30m/s
So, the rate at which the avalanche is moving from 0s to 10s is 30m/s.
c. To find the rate at which the avalanche is moving at 10s, we need to find the derivative of the equation with respect to t:
s'(t) = d(s(t))/dt = d(3t^2)/dt = 6t
Substituting t = 10s into the derivative:
s'(10) = 6(10)
s'(10) = 60m/s
So, the rate at which the avalanche is moving at 10s is 60m/s.
d. To find how long it takes for the leading edge of the snow to move 600m, we need to solve the equation:
600 = 3t^2
Dividing both sides by 3:
200 = t^2
Taking the square root of both sides:
t = sqrt(200)
t ≈ 14.14s
To the nearest second, the leading edge of the snow takes approximately 14 seconds to move 600m.
To find the answers to the given questions, we first need to understand the equation that describes the motion of the avalanche: s(t) = 3t^2, where s is the distance in meters traveled by the leading edge of the snow at t seconds.
a. To find the distance traveled from 0s to 5s (s(5) - s(0)):
Substitute t = 5 into the equation: s(5) = 3(5^2) = 3(25) = 75 meters.
Therefore, the distance traveled from 0s to 5s is 75 meters.
b. To find the rate at which the avalanche is moving from 0s to 10s, we need to find the derivative of the equation with respect to time (t):
s'(t) = d/dt (3t^2) = 6t.
Substitute t = 10 into the derivative equation: s'(10) = 6(10) = 60 m/s.
Therefore, the rate at which the avalanche is moving from 0s to 10s is 60 m/s.
c. To find the rate at which the avalanche is moving at 10s, we also use the derivative equation: s'(t) = 6t.
Substitute t = 10 into the derivative equation: s'(10) = 6(10) = 60 m/s.
Therefore, the rate at which the avalanche is moving at 10s is 60 m/s.
d. To find how long, to the nearest second, it takes for the leading edge of the snow to move 600m, set up the equation s(t) = 600 and solve for t:
3t^2 = 600.
Divide both sides by 3: t^2 = 200.
Take the square root of both sides: t = √200.
To the nearest second, t is approximately 14 seconds.
Therefore, the leading edge of the snow takes approximately 14 seconds to move 600 meters.