# Math Help - can someone help with this please

1. ## can someone help with this please

The angular displacement, qradians, of the spoke of a wheel is given by
the expression q= 0.5t4 t3, where t is time in seconds. Find:

(a) The angular velocity after 2 seconds

(b) The angular acceleration after 3 seconds

(c) The time when the angular acceleration is zero

2. ## Re: can someone help with this please

Originally Posted by tomjay
The angular displacement, qradians, of the spoke of a wheel is given by
the expression q= 0.5t4 t3, where t is time in seconds. Find:

(a) The angular velocity after 2 seconds

(b) The angular acceleration after 3 seconds

(c) The time when the angular acceleration is zero
Velocity is the derivative of displacement, and acceleration is the derivative of velocity.

3. ## Re: can someone help with this please

a)differentiate the expression once and substitute t=2.
b)differentiate the expression twice and substitute t=3
c) Equate the twice differentiated expression to zero and find 't'.

4. ## Re: can someone help with this please

What ive done
a) Velocity = d Ɵ /dt

=2t^3 – 3t^2 = V

at t = 2,

velocity = 2(2^3) – 3(2^2)
=16 – 12

b) Acceleration = dv/dt

= 6t^2 – 6t =A

At t = 3

Acceleration = 6 (3^2) – 6 (3)
= 54 – 18

c) A = dv/dt = 6t^2 – 6t = 0

6t^2 = 6t

Acceleration = 0 when t = 1 second

5. ## Re: can someone help with this please

correct...Did you ask a doubt or r u testing our knowledge?

6. ## Re: can someone help with this please

Originally Posted by tomjay
What ive done
a) Velocity = d Ɵ /dt

=2t^3 – 3t^2 = V

at t = 2,

velocity = 2(2^3) – 3(2^2)
=16 – 12

b) Acceleration = dv/dt

= 6t^2 – 6t =A

At t = 3

Acceleration = 6 (3^2) – 6 (3)
= 54 – 18

c) A = dv/dt = 6t^2 – 6t = 0

6t^2 = 6t

Acceleration = 0 when t = 1 second
Always solve quadratics by factorising or using the Quadratic Formula/Completing the Square. By dividing both sides by t, you are assuming that \displaystyle \begin{align*} t \neq 0 \end{align*}.

\displaystyle \begin{align*} 6t^2 - 6t &= 0 \\ 6t(t - 1) &= 0 \\ 6t = 0 \textrm{ or } t- 1 &= 0 \\ t = 0 \textrm{ or } t &= 1 \end{align*}

There is also 0 acceleration when t = 0.

7. ## Re: can someone help with this please

Originally Posted by tomjay
What ive done
a) Velocity = d Ɵ /dt

=2t^3 – 3t^2 = V

at t = 2,

velocity = 2(2^3) – 3(2^2)
=16 – 12
Yes, that is correct.

b) Acceleration = dv/dt

= 6t^2 – 6t =A

At t = 3

Acceleration = 6 (3^2) – 6 (3)
= 54 – 18