Question

a dog initially running at a speed of 2m/s, accelerates at a constant rate of 1.5 m/s² for 5 s. calculate (1) distance covered by the dog from its initial position (2) the final velocity of the dog .​

Answer 1

Answer:

(1) 28.75 m

(2) 9.5 m/s

Explanation:

Given :

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

To find :

• the distance covered by the dog from its initial position
• the final velocity of the dog

Solution :

initial speed, u = 2 m/s

acceleration, a = 1.5 m/s²

time taken, t = 5 s

From the second equation of motion, the distance covered by an object in t sec is given by,

$\boxed{\tt S=ut+ \dfrac{1}{2}at^2}$

Substitute the values,

S = 2(5) + ½(1.5)(5)²

S = 10 + ½(1.5)(25)

S = 10 + 18.75

S = 28.75 m

Therefore, the distance covered by the dog from initial position is 28.75 m

From the first equation of motion, final velocity is given by;

v = u + at

Substitute the values,

v = 2 + (1.5)(5)

v = 2 + 7.5

v = 9.5 m/s

The final velocity is 9.5 m/s

Answer 2

Answer:

Given :-

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

To Find :-

1. Distance covered by the dog from its initial position.
2. The final velocity of the dog.

Formula Used :-

➲ To find distance covered we know that,

$\longmapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}a{t}^{2}}}}$

where,

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

➲ To find final velocity we know that,

$\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

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

Solution :-

1) Distance covered by the dog from its initial position :

Given :

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

According to the question by using the formula we get,

↦ $\sf s =\: 2 \times 5 + \dfrac{1}{2} \times 1.5 \times {(5)}^{2}$

↦ $\sf s =\: 10 + \dfrac{1}{2} \times 1.5 \times 25$

↦ $\sf s =\: 10 + \dfrac{1}{2} \times 37.5$

↦ $\sf s =\: 10 + \dfrac{37.5}{2}$

↦ $\sf s =\: 10 + 18.75$

➠ $\sf\bold{\red{s =\: 28.75\: m}}$

$\therefore$ The distance covered by the dog from its initial position is 28.75 m .

$\rule{300}{2}$

2) The final velocity of the dog :

Given :

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

According to the question by using the formula we get,

↦ $\sf v =\: 2 + 1.5 \times 5$

↦ $\sf v =\: 2 + 7.5$

➠ $\sf\bold{\red{v =\: 9.5 \: m/s}}$

$\therefore$ The final velocity of the dog is 9.5 m/s.