Question

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
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Answer 1

Answer:

i)Here, u=2 m/s

a=1.5m/s^2

t=5s

Then,s=ut+1/2 at^2

=2*5+1/2*1.5*(5^2) m

=10+1/2*37.5 m

=10+18.75 m

=28.75 m

ii) Now, s=28.75 m

Then, v^2=u^2+2as

or, v^2=2^2+2*1.5*5 m/s

or, v^2=4+15 m/s

or, v= ¬19 m/s

so, v=4.36 m/s

Explanation:

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Answer 2

Given :

• Initial velocity (u) = 2 m/s
• Acceleration (a) = 1.5 m/s²
• Time (t) = 5 seconds

To find :

• Distance (s)
• Final velocity (v)

According to the question,

i)

By using Newtons second equation of motion,

→ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• t = Time
• a = Acceleration

→ s = 2 × 5 + ½ × 1.5 × 5 × 5

→ s = 10 + ½ × 37.5

→ s = 10 + 18.75

→ s = 28.75

So,the distance covered by the dog is 28.75 metre.

ii)

By using Newtons first equation of motion,

→ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• t = Time
• a = Acceleration

→ v = 2 + 1.5 × 5

→ v = 2 + 7.5

→ v = 9.5

So,the final velocity is 9.5 m/s.