Question

A cycle at rest is acted upon by a force and it aquires 10 m/sec velocity in 5 seconds.Calculate the distance travelled by the body.

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 0 m/s

Final velocity, v = 10 m/s

Time taken, t = 5 econds

To Find,

Distance covered, s = ?

1st and 2nd equations of motion,

v = u + at and s = ut + 1/2at²

v = u + at

⇒ 10 = 0 + a × 5

⇒ 10 = 5a

⇒ 10/5 = a

⇒ a = 2 m/s²

Here, the acceleration is 2 m/s².

Now, the distance covered,

s = ut + 1/2at²

⇒ s = 0 × 5 + 1/2 × 2 × (5)²

⇒ s = 1/2 × 2 × 25

⇒ s = 25 m

Hence, the distance covered is 25 m.

Answer 2

Given ,

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 10 m/s
• Time (t) = 5 sec

We know that , the Newton's first equation of motion is given by

$\large \boxed{ \sf{v = u + at}}$

Thus ,

10 = 0 + 5a

a = 10/5

a = 2 m/s²

Now , the Newton's second equation of motion is given by

$\large \boxed{ \sf{s = ut + \frac{1}{2} a {(t)}^{2} }}$

Thus ,

$\sf \mapsto s = 0 \times 5 + \frac{1}{2} \times 2 \times {(5)}^{2} \\ \\ \sf \mapsto s = {(5)}^{2} \\ \\ \sf \mapsto s = 25 \: \: m$

$\sf \therefore\underline{The \: distance \: travelled \: by \: the \: body \: is \: 25 \: m}$