Aman hits a ball and provides it with an initial velocity of 5.0 m/s on a rough horizontal surface. due to the surface the ball slows down with an acceleration of 0.5 m/s^2 . when the ball is hit, a dog wihch is 1.0 m behind the man runs with acceleration 1.5 m/s^2 starting from the rest to catch the ball.

a) when does the dog catch the ball?

b)how far away from the dog's initial position, does the dog catch the ball?

c)what is the speed of the ball and the dog when the dog catchesthe ball?

a) After 5.19 seconds dog catch the ball.

b)  From the dog's initial position at 20.20 m dog catches the ball.

c) Speed of the ball when dog catches ball = 2.405 m/s

   Speed of the dog when dog catches ball = 7.785 m/s

Explanation:

a) Let the time of catching be t.

   We have equation of motion s = ut + 0.5 at²

   Consider the motion of ball

                Initial velocity, u = 5 m/s

                Acceleration, a = -0.5 m/s²

                Time, t = t

                Substituting

                 s = 5 x t + 0.5 x -0.5 x t²

                 s = 5t - 0.25t²

  Consider the motion of dog

                 Initial velocity, u = 0 m/s

                Acceleration, a = 1.5 m/s²

                Time, t = t

                Substituting

                 s + 1 = 0 x t + 0.5 x 1.5 x t²

                 s = 0.75t²      

If they catch up displacement of dog is 1 m more than displacement of ball.

That is

                5t - 0.25t² + 1 =   0.75t²  

                t² - 5t -1 = 0

                t = 5.19 or t = -0.19(not possible)

So after 5.19 seconds dog catch the ball.

b) Displacement of dog, s = 0.75t²  

                            s = 0.75 x 5.19²

                             s = 20.20 m

    From the dog's initial position at 20.20 m dog catches the ball.

c) We have equation of motion v = u + at

       Consider the motion of ball

                Initial velocity, u = 5 m/s

                Acceleration, a = -0.5 m/s²

                Time, t = 5.19 s

                Substituting

                           v = 5 + -0.5 x 5.19 = 2.405 m/s

                Speed of the ball when dog catches = 2.405 m/s

  Consider the motion of dog

                Initial velocity, u = 0 m/s

                Acceleration, a = 1.5 m/s²

                Time, t = 5.19 s

                Substituting

                           v = 0 + 1.5 x 5.19 = 7.785 m/s

                Speed of the dog when dog catches ball = 7.785 m/s

higher potential

explanation:

because it is in a high speed so it must move into a of high potential

chemical energy --> gravitational potential energy --> kinetic energy --> elastic potential energy --> gravitational potential energy

explanation:

when emma starts the hike, the chemical energy stored in her body is used to climb up the mountain, and so she gains gravitational potential energy (because at the end, she will be at a higher height h above the ground). then she goes bungee jumping: as she falls, this gravitational potential energy is converted into kinetic energy (because her speed increases as she falls down), until the point when the rope stretches, and she stops moving. at that moment, all the kinetic energy is converted into elastic potential energy due to the stretching of the rope. then, this energy is released because the rope returns to its natural length, and emma gains a small increase in altitude , so she gains some gravitational potential energy.