Question

Practice Problems 3
1 A dog initially running at a speed of 2 m/s, accelerates at a constant rate of 1.5 m/s2 for 5 s.
Calculate (i) the distance covered by the dog from its initial position, (ii) the final velocity of
the dog
[(i) 28.75 m (ü) 9.5 m/s​

Answer 1

Explanation:

Here, the values of initial velocity (u), acceleration (a) and time (t) are already given. So by using the above formulaes, we can find the values of final velocity and the distance covered by the dog.

( I've attached the solution)

Answer 2

Given data :-

Here,

{ u = initial velocity, v = final velocity, a = acceleration of the particle, t = time taken by particle, s = displacement of the particle. }

• A dog initially running at a speed of 2 m/s, accelerates at a constant rate of 1.5 m/s² for 5 s.

→ initial velocity of the dog ( u ) = 2 m/s

→ final velocity of the dog ( v ) = ?

→ acceleration of the dog ( a ) = 1.5 m/s²

→ time taken by the dog ( t ) = 5 sec

→ displacement of the dog ( s ) = ?

Solution : -

Here we use kinematical equation to find distance covered by the dog from its initial position

→ s = ut + ½ at²

→ s = 2 × 5 + ½ × 1.5 × ( 5 )²

→ s = 10 + ½ × 1.5 × 25

→ s = 10 + ½ × 37.5

→ s = 10 + 18.75

→ s = 28.75 m

Now to find final velocity of dog

→ v = u + at

→ v = 2 + 1.5 × 5

→ v = 2 + 7.5

→ v = 9.5 m/s

Hence,

( i ) The distance covered by the dog from its initial position is 28.75 m.

( ii ) The final velocity of dog is 9.5 m/s