What is the velocity of the object when the acceleration is 0 (with given equation)?

A. Acceleration is D ''(t)=6t-30 meters. When D ''(t) = 0, then we can find t=5.

Now, D'(t) = 3t^2-30t+80 and we can plugin 5 for t to get D'(t) = -5

B. Velocity = 0 when 3t^2-30t+80 = 0, which happens when t = -80 or t = 10

I'm going to assume you know that acceleration is the derivative of velocity, and velocity is the derivative of distance. By taking the derivatives, you get

D(t) = t^3 - 15t^2 + 80t

V(t) = 3t^2 - 30t + 80

A(t) = 6t -30

Therefore,

A(t) = 0 = 6t - 30

t = 5 [s]

So,

V(5) = 3(5)^2 - 30(5) +80

V = 5 [m/s]

Since velocity is a quadratic, you can just graph it to see if it has zeros or you can try to find its zeros manually. There are no zeros in this case, and you can show this using the quadratic formula.

a. d(t) = t^3 - 15t^2 + 80t

Take the derivative to find the velocity.

v(t) = 3t^2 - 30t + 80

Take the derivative again to find the acceleration.

a(t) = 6t - 30

Set this (a(t)) equal to 0 to find t.

0 = 6t - 30

6t = 30

t = 5

Plug t = 5 into the expression for velocity;

v(5) = 3(5)^2 - 30(5) + 80

v(5) = 5

The velocity is 5 meters/second.

b. Set the expression for velocity equal to 0, then solve for t.

0 = 3t^2 - 30t + 80

Because this is a quadratic, take the discriminant to test for real solutions:

b^2 - 4ac = (-30)^2 - 4(3)(80) = -60

Since the discriminant is negative, there are no real solutions. Thus, the velocity is never zero.

Hope this helps!

That equation denotes the objects position; take the derivative of it to obtain the objects velocity equation; and once again for acceleration.

D'(t) = 3t^2-30t+80 (velocity)

D''(t) = 6t-30 (acceleration)

a) Plug in 0 to the acceleration equation.

0 = 6t-30

30 = 6t

t = 5 (time when acceleration is 0)

Plug in 5 into the velocity equation for time.

D'(t) = 3(5)^2-30(5)+80

D'(t) = 75-150+80

D'(t) = 5 m/s

ANSWER: 5 m/s

b) Plug in 0 into the velocity equation.

0 = 3t^2-30t+80

Notice that this equation has potential to be "foiled" or be put through the quadratic equation. If you get an real values (t greater than or equal to 0) then the answer is yes, if not then no.

d = t^3-15t^2+80t

v= 3t^2-30t+80

a=6t-30=0 => t=5

v(5) = 3* 25 -30 * 5 + 80

= 75+80-150 = 155-150 = 5 m/s

3t^2-30t+80 = 3 (t^2-10+80/3) = 3[(t -5 )^2 -25+80/3]

= 3[(t-5)^2-25 + 26 2/3] = 3[(t-5)^2 +5/3]

so no the velocity is never zero

the main person-friendly SUVAT equation of all of them is distance = common velocity X time traveled. In math communicate that's S = Vavg T. and that's the place you could initiate quite lots each and all the time whilst doing kinematics. Assuming A = consistent acceleration, in spite of that's, we are able to instruct that Vavg = (V + U)/2 the place V is end velocity and U is preliminary velocity. So Vavg = S/T = (V + U)/2 and remedy for V. 2S = (V + U)T so as that 2S - UT = VT and V = 2S/T - U and there you're: distance (S), time (T), and preliminary velocity (U). QED. Your difficulty is the form you're reading physics. you would be gaining information of the physics, e.g., distance = common velocity X time traveled incredibly than memorizing those equations that have served you no purpose. Had you started with the physics and derived the equation, as I did right here, you have gotten responded this incredibly.

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