700cos6=-100cos45° 17005i=-100545° 8=arccos()=arccos=84.2 vma=700sin6+100sin45°=767.1m/s Thus the direction which the plane must be aimed is West to North 84.2" or North to West 5.8. The ground speed of the plane is 767. Im/s 4. A particle moving clockwise in a circle with a radius of 3. 00 m has a total acceleration at some instant of magnitude 15.0 m/s- directed as indicated in Figure 4 (a) Find the magnitude of the centripetal acceleration at this instant (b) Find the speed of the particle at this instant. (c)Find the angular speed of the particle at this instant (d) Determine the magnitude of the tangential acceleration at this instant (e) determine the magnitude of the angular acceleration at this instant (f) Make a sketch indicating the directions of @, a, r and v ( g) Is the particle speeding up or slowing down in its circular motion? Solution: (a)The magnitude of the centripetal acceleration at this instant is an= a cOS40°=15cos40°=1.5(m/s2) (b)Using We have the speed of the particle at this instant v=var=5.87(m/s) (c)The angular speed of the particle at this instant is @===1-C=1.96 rad/s (d) The magnitude of the tangential acceleration at this instant a1= a sin40°=15sin40°=96(m/s2) (e)Using d,=a×F,a1=ar We have the magnitude of the angular acceleration at this instant a=t=3.2(rad/s2) (f The directions of a, a, r and v shown in figure (g) Because the angle between a and v is 60%, it is less than T/2, so the particle is speeding⎩ ⎨ ⎧ = − − = − ⇒ o o 700sin 100sin 45 700cos 100cos45 PG θ v θ ⎪ ⎩ ⎪ ⎨ ⎧ = + = = = = ⇒ 700sin 100sin 45 767.1m/s 84.2 14 2 ) arccos 14 2 arccos( o o θ θ PG v Thus the direction which the plane must be aimed is West to North 84.2˚ or North to West 5.8˚. The ground speed of the plane is 767.1m/s. 4. A particle moving clockwise in a circle with a radius of 3.00 m has a total acceleration at some instant of magnitude 15.0 m/s2 directed as indicated in Figure 4. (a) Find the magnitude of the centripetal acceleration at this instant. (b) Find the speed of the particle at this instant. (c) Find the angular speed of the particle at this instant. (d) Determine the magnitude of the tangential acceleration at this instant. (e) Determine the magnitude of the angular acceleration at this instant. (f) Make a sketch indicating the directions of ω r , α r , r r and v r . (g) Is the particle speeding up or slowing down in its circular motion? Solution: (a) The magnitude of the centripetal acceleration at this instant is ac = atotal cos40° =15cos40° =11.5 (m/s2 ) (b) Using r v ac 2 = , We have the speed of the particle at this instant v = ac r = 5.87 (m/s) (c) The angular speed of the particle at this instant is = = = 1.96 r a r v c ω rad/s (d) The magnitude of the tangential acceleration at this instant is at = atotalsin 40° =15sin 40° = 9.6 (m/s2 ) (e) Using a r a r t = α × , t = α r r r We have the magnitude of the angular acceleration at this instant = = 3.2 r at α (rad/s2 ) (f) The directions of ω r , α r , r r and v r shown in figure. (g) Because the angle between a v and v v is 60˚, it is less than π / 2 , so the particle is speeding up. 40.0° atotal r Fig.4 v r v v a v ω v