A jet plane is flying at a constant altitude. At time t1 = 0, it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t2 = 30.0 s, the components are vx = - 170 m/s, vy = 40 m/s. (a) Sketch the velocity vectors at t1 and t2. How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.
Solution 5E Step 1 : In this question, we need to Sketch velocity vectors for time t1 and t2 And find the difference between these two vectors Find the components of average acceleration of jet plane Find the magnitude and direction of the average acceleration Considering the data given We have Velocity along x-component at time t1=0s v x1= 90 m/s Velocity along y-component at time t1=0s v y1= 110 m/s Velocity along x-component at time t2=30s v x2 = 170 m/s Velocity along y-component at time 2t=30s v y2 = 40 m/s Step 2 : The velocity vectors for v xnd v ay time t= 0 and t=30s We have Step 3 : Let us find the acceleration of the jet plane We know that acceleration is given by change in velocity with respect to time, hence we can write a = dv dt but we already have values for v and vxwhich aye the velocity component along x and y axis respectively And the time interval is given by t = t`2 t 1 30s 0s = 30 s Let us find the change in velocity along x-component v = v v x x2 x1 Substituting the values we have vx= 170 m/s 90 m/s = 260 m/s Hence we have v = 80xm/s Similarly Let us find the change in velocity along y-component vy= v y2 y1 Substituting the values we have vy= 40 m/s 110 m/s = 70 m/s Hence we have v = y0 m/s Now the acceleration along x - component is given by ax= vx t Substituting the values we get ax= 260 m/s 30 s 2 ax= 8.66 m/s Hence we have obtained acceleration along x- component as 8.66 m/s 2 Similarly The acceleration along y - component is given by a = vy y t Substituting the values we get 70 m/s a = 30 s ay= 2.33 m/s 2 2 Hence we have obtained acceleration along y- component as 2.33 m/s