a particle moves in a straight line so that t seconds after passing a fixed point O its acceleration a cm per second is given by a=4t-16. It passes O with a velocity of 30cm per second.
Find an expression for the velocity of the psrticle in terms of t and show that it comes to instantenous rest when t=3
a = 4t - 16
integrating ...
v = 2t^2 - 16t + c
given: at t=0, v = 30
30 = 0-0+c
c = 0
velocity = 2t^2 - 16t + 30
when t=3
velocity = 2(9) - 16(3) + 30 = 0
looks like it is resting.
Well, I must say, this particle seems to be quite the daredevil, zooming around with all that acceleration and velocity. Let's see if we can make sense of it all.
To find the expression for the velocity of the particle in terms of time, we need to integrate the acceleration function. So, let's break out our integration cap and get to work.
∫(4t - 16) dt = 2t^2 - 16t + C
Where C is the constant of integration.
Now, we know that the particle passes point O with a velocity of 30 cm per second. So let's plug in t = 0 (since it's right after passing O) and solve for C.
30 = 2(0)^2 - 16(0) + C
C = 30
Therefore, the expression for the velocity of the particle is:
v(t) = 2t^2 - 16t + 30
Now, to address the claim that the particle comes to instantaneous rest when t = 3. Well, let's plug in t = 3 and see what happens.
v(3) = 2(3)^2 - 16(3) + 30
v(3) = 18 - 48 + 30
v(3) = 0
Well, well, well! It seems like our dear particle indeed comes to instantaneous rest when t = 3. I guess even particles need a little break sometimes.
Keep those questions coming, and I'll keep the humor flowing!
To find an expression for the velocity of the particle in terms of time, we need to integrate the acceleration with respect to time.
Given that the acceleration is given by a = 4t - 16, we can integrate this equation to find the velocity, v.
∫a dt = ∫(4t - 16) dt
Integrating, we get:
v = 2t^2 - 16t + C
where C is the constant of integration.
Next, we are given that the particle passes point O with a velocity of 30 cm/s. We can use this information to find the value of C.
When t = 0, v = 30:
30 = 2(0)^2 - 16(0) + C
Simplifying this equation, we find:
C = 30
Therefore, the expression for the velocity of the particle in terms of t is:
v = 2t^2 - 16t + 30
To find when the particle comes to instantaneous rest, we set v = 0:
0 = 2t^2 - 16t + 30
This is a quadratic equation. Solving for t, we find:
t = 3
Therefore, the particle comes to instantaneous rest when t = 3.
To find an expression for the velocity of the particle in terms of time (t), we can integrate the acceleration (a) with respect to time.
Given:
Acceleration (a) = 4t - 16
To find the velocity (v), we integrate the acceleration over the time interval from 0 to t:
∫a dt = ∫(4t - 16) dt
Integrating 4t with respect to t gives us 2t^2, and integrating -16 gives us -16t:
v = ∫(4t - 16) dt = 2t^2 - 16t + C
Now we need to find the constant of integration (C). We're given that the particle passes point O with a velocity of 30 cm/s. This means that when t = 0, v = 30:
30 = 2(0)^2 - 16(0) + C
30 = C
Therefore, the expression for the velocity of the particle in terms of t is:
v = 2t^2 - 16t + 30
To show that the particle comes to instantaneous rest when t = 3, we substitute t = 3 into the expression for velocity:
v = 2(3)^2 - 16(3) + 30
v = 18 - 48 + 30
v = 0
When t = 3, the velocity of the particle is 0 cm/s. Therefore, the particle comes to instantaneous rest at t = 3.