Question

10. An automobile, traveling at a constant speed of 25 m/s, enters a 90° curve and emerges from this curve 68 later. What are the x and y components of average acceleration during this time interval? [Hint: Av, 0 m/s - 25 m/s -25 m/s, Avy - 25 m/s - 0 m/s - 25 m/s.az - Av, /t, a, Av, /t] 11. While jogging this morning you encountered a circular track. As you finished one round along the track with uniform speed, what were your average velocity and average acceleration for this motion? 12. A motorcycle is traveling clockwise around a traffic circle at a steady speed of 50 km/h. The x axis is considered eastward and y axis is northward. What are the x and y components of velocity at a point when the motorcycle is at the eastern point of the circle? (Hint: velocity is along the tangent at any point) 13. A tennis ball, initially traveling horizontally with a velocity of 13 m/s, collides with a wall and bounces back with the same speed. If the collision lasts for 0.4 s, what is the magnitude and direction of the acceleration during the collision? [Ans: 65 m/s in the direction of reversed motion]

Answer 1

10.

10 som Giney here a Jo Aveo - 25m/= -25 m/s darr in by = -25oms 1-o m/s = -2mls t= 65 e ak = AM -25 -4.168 mg-2 Vy ... & ay= Ave. - 28. - 4. 166omst x and r component of average acceleration during the tone internal of 6 see are ax= -4.166 mse &ag= – 4.166 m52 +

11.The average acceleration will be a zero, since in one complete round the displacement is zero which implies that average velocity is also zero which further implies that average acceleration will also be azero for a me when I having a uniform circular motion with uniform speed.

12. When motorcycle will be at eastern part of the circle then the

X component of velocity will be Vx = 0

Y component of Velocity will be Vy = 50 km/ h

As at every point the velocity will be tangential to the circle.

13.

Acceleration is given as = rate of change if velocity with respect to time

So there acceleration = (V2-V1)/t

a = (-13-(13))/0.4 = -  65 m/s2

So that magnitude of acceleration will be 65m/s2 and here negative sign show that acceleration is in reverse direction.

As the direction of the velocity after the collision is reversed so that i have taken it in negative direction.