A particle moves in accordance with the equation s=32+24t-2t?
Find a) v-t relationship
b) How far to the right of the origin does the particle go?
c) s when t=0
d) acceleration of the particle at t=3sec
a) v-t relationship: v = 24 - 2t
b) How far to the right of the origin does the particle go? 32 units
c) s when t=0: 32
d) acceleration of the particle at t=3sec: -2 m/s^2
Let's solve the given equations step-by-step:
a) To find the velocity-time relationship, we differentiate the equation s = 32 + 24t - 2t^2 with respect to time (t). Differentiating s with respect to t gives us the velocity (v).
ds/dt = 24 - 4t
So, the velocity-time relationship is v = 24 - 4t.
b) To find how far to the right of the origin the particle goes, we need to find the displacement when the particle returns to the origin. The displacement is given by the s equation when it equals zero.
0 = 32 + 24t - 2t^2
To solve for t, we can rearrange the equation into standard quadratic form:
2t^2 - 24t - 32 = 0
Now let's solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = -24, and c = -32:
t = (-(-24) ± √((-24)^2 - 4(2)(-32))) / (2 * 2)
t = (24 ± √(576 + 256)) / 4
t = (24 ± √832) / 4
t ≈ (24 ± 28.86) / 4
Since we are looking for the time when the particle returns to the origin, we can ignore the negative solution:
t ≈ (24 + 28.86) / 4 ≈ 13.72/4 ≈ 3.43 seconds
Therefore, the particle travels to the right of the origin for approximately 3.43 seconds.
c) To find s when t = 0, we substitute t = 0 into the equation s = 32 + 24t - 2t^2:
s = 32 + 24(0) - 2(0)^2
s = 32
Therefore, when t = 0, s = 32.
d) To find the acceleration of the particle at t = 3 seconds, we differentiate the velocity equation v = 24 - 4t with respect to time (t), which will give us the acceleration (a).
dv/dt = -4
Therefore, the acceleration of the particle at t = 3 seconds is a = -4.
To find the answers to these questions, we need to understand the given equation and the relationships between position (s), velocity (v), and acceleration (a).
The equation given is: s = 32 + 24t - 2t^2
a) To find the v-t relationship, we need to differentiate the equation with respect to time (t). Differentiating s with respect to t gives us velocity (v). Let's differentiate the equation:
ds/dt = d(32 + 24t - 2t^2)/dt
= 24 - 4t
Therefore, the v-t relationship is: v = 24 - 4t.
b) To find how far to the right of the origin the particle goes, we need to find the position (s) when the velocity (v) is zero. This occurs when 24 - 4t = 0. Solving for t:
24 - 4t = 0
4t = 24
t = 6
We substitute t=6 into the equation for s:
s = 32 + 24(6) - 2(6)^2
= 32 + 144 - 72
= 104
Therefore, the particle goes 104 units to the right of the origin.
c) The position (s) when t = 0 can be found by substituting t = 0 into the equation for s:
s = 32 + 24(0) - 2(0)^2
= 32
Therefore, when t = 0, the position of the particle is 32 units.
d) To find the acceleration (a) of the particle at t = 3 seconds, we need to differentiate the velocity equation (v) with respect to time (t). Let's differentiate v:
dv/dt = d(24 - 4t)/dt
= -4
Therefore, the acceleration of the particle at t = 3 seconds is -4 m/s^2.