A dog initial running at a speed of 2m/s, accelates at a constant rate of 1.5 m/s² for 5 s. Calculate (i) the distance cover by the dog from its initial position, (ii) the final velocity of the dog.
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Given data :-
- A dog initial running at a speed of 2 m/s, accelates at a constant rate of 1.5 m/s² for 5 sec.
To find :-
- The distance cover by the dog from its initial position.
- The final velocity of the dog.
Solution :-
Here,
→ u = initial velocity of the dog
→ v = final velocity of the dog
→ a = uniform acceleration of the dog
→ s = displacement of the dog
→ t = time taken by dog to cover distance
According to given
→ u = 2 m/s
→ a = 1.5 m/s²
→ t = 5 sec
Now,{ we use kinematical equation to find final velocity of dog and distance cover by dog }
→ 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,
→ v = u + at
→ v = 2 + 1.5 × 5
→ v = 2 + 7.5
→ v = 9.5 m/s
Hence, the distance cover by the dog from its initial position is 28.75 m and the final velocity of the dog is 9.5 m/s.
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