Results 1 to 7 of 7

- August 11th 2012, 01:13 AM #1

- Joined
- Jun 2012
- From
- UK
- Posts
- 39

## can someone help with this please

The angular displacement, qradians, of the spoke of a wheel is given by

the expression q= 0.5*t*^{4 }–*t*^{3}, 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

- August 11th 2012, 01:47 AM #2

- August 11th 2012, 02:16 AM #3

- Joined
- Aug 2012
- From
- banglore
- Posts
- 4

- August 11th 2012, 02:24 AM #4

- Joined
- Jun 2012
- From
- UK
- Posts
- 39

## 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

= 4 Radians/second

**b)**Acceleration = dv/dt

= 6t^2 – 6t =A

At t = 3

Acceleration = 6 (3^2) – 6 (3)

= 54 – 18

36 Radians/second^2

**c)**A = dv/dt = 6t^2 – 6t = 0

6t^2 = 6t

Acceleration = 0 when t = 1 second

- August 11th 2012, 02:43 AM #5

- Joined
- Aug 2012
- From
- banglore
- Posts
- 4

- August 11th 2012, 02:49 AM #6

- August 11th 2012, 06:10 AM #7

- Joined
- Apr 2005
- Posts
- 17,727
- Thanks
- 2270

## Re: can someone help with this please

Yes, that is correct.

**b)**Acceleration = dv/dt

= 6t^2 – 6t =A

At t = 3

Acceleration = 6 (3^2) – 6 (3)

= 54 – 18

36 Radians/second^2

**c)**A = dv/dt = 6t^2 – 6t = 0

6t^2 = 6t

Acceleration = 0 when t = 1 second**quadratic**equation and has**two**solutions. Yes, one solution is t= 1. What is the other?