A man 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