The Displacement (in metres) of an object moving in a straight line is given by s=1+ 2t + 1/4 t^2, where t is measures in seconds.
a) Find the average velocity over the time interval [1,2]. do not interpret the physical meaning of the value for this question.
b)Find the instantaneous velocity when t =1. do not interpret the physical meaning of the value for this question.
(a) (s(2)-s(1))/(2-1)
(b) v = ds/dt = 2 + 1/2 t ... evaluate at t=1
a) Well, let's get calculating the average velocity! To find the average velocity over the interval [1,2], we need to find the displacement at both t=1 and t=2 first.
At t=1, the displacement is given by s=1+2(1) + 1/4(1)^2, which is equal to 1 + 2 + 1/4, making it 2.25 meters.
At t=2, the displacement is given by s=1+2(2) + 1/4(2)^2, which simplifies to 1 + 4 + 1/4(4), resulting in 6.25 meters.
Now that we have the displacements, we can find the average velocity by taking the difference in displacements and dividing it by the difference in time. So, the average velocity is (6.25 - 2.25) / (2 - 1), giving us a value of 4 meters per second.
b) Oh, instantaneous velocity! Fancy! To find that, we need to take the derivative of the displacement equation with respect to time. But don't worry, I'll do the math for you.
The derivative of s=1+2t+1/4t^2 with respect to t is v=2+(1/2)t.
Now, we can substitute t=1 into the equation to find the instantaneous velocity. Plugging in t=1, we get v=2+(1/2)(1), which gives us a value of 2.5 meters per second.
a) To find the average velocity over the time interval [1,2], we need to calculate the displacement over that interval and divide it by the time taken.
First, let's find the displacement at t = 1:
s(1) = 1 + 2(1) + (1/4)(1)^2
s(1) = 1 + 2 + 1/4
s(1) = 3.25
Next, let's find the displacement at t = 2:
s(2) = 1 + 2(2) + (1/4)(2)^2
s(2) = 1 + 4 + 1/4(4)
s(2) = 6.5
Now, we can calculate the average velocity using the formula:
Average velocity = (displacement) / (time)
Average velocity = (s(2) - s(1)) / (2 - 1)
Average velocity = (6.5 - 3.25) / 1
Average velocity = 3.25
Therefore, the average velocity over the time interval [1,2] is 3.25.
b) To find the instantaneous velocity when t = 1, we need to find the derivative of the displacement function with respect to time and evaluate it at t = 1.
Taking the derivative of the displacement function s(t) = 1 + 2t + (1/4)t^2:
s'(t) = 0 + 2 + (1/2)(t)
s'(t) = 2 + (1/2)t
Now, let's evaluate s'(t) at t = 1:
s'(1) = 2 + (1/2)(1)
s'(1) = 2 + 1/2
s'(1) = 2.5
Therefore, the instantaneous velocity when t = 1 is 2.5.
a) To find the average velocity over the time interval [1,2], we need to calculate the displacement of the object at both times and then divide the difference in displacement by the difference in time.
Given the equation for displacement as s = 1 + 2t + 1/4 t^2, we can find the displacement at t = 1 and t = 2 by substituting these values into the equation:
s(1) = 1 + 2(1) + 1/4(1)^2 = 1 + 2 + 1/4 = 9/4
s(2) = 1 + 2(2) + 1/4(1)^2 = 1 + 4 + 1/4 = 13/4
Now we can calculate the average velocity:
Average velocity = (s(2) - s(1)) / (2 - 1)
= (13/4 - 9/4) / (2 - 1)
= 4/4
= 1
Therefore, the average velocity over the time interval [1,2] is 1.
b) To find the instantaneous velocity when t = 1, we need to take the derivative of the displacement equation with respect to time. The resulting derivative will give us the instantaneous velocity at any given time.
Given the equation for displacement as s = 1 + 2t + 1/4 t^2, we differentiate it with respect to t:
ds/dt = d/dt(1) + d/dt(2t) + d/dt(1/4 t^2)
= 0 + 2 + (1/4)(2t)
= 2 + (1/2)t
Now we can substitute t = 1 into the derivative equation to find the instantaneous velocity:
v(1) = 2 + (1/2)(1)
= 2 + 1/2
= 5/2
Therefore, the instantaneous velocity when t = 1 is 5/2.