Question

A body starts from rest and accelerates for 5m/s. Calculate the time taken by the body to cover a distance of 1km

Answer 1

Answer:

Time taken is 20 seconds.

Explanation:

To find :

• Time taken by the body.

We have,

• If body starts from rest. Then, Initial velocity, u is 0 m/s.
• Acceleration, a is 5 m/s²
• Distance, s is 1 km.

Conversion :

• Distance = 1 km

1 km = 1000 m

So, Distance, s is 1000 m .

We know,

Second equation of motion :

• s = ut + 1/2 at²

[Where, s is distance, u is Initial velocity, t is time taken and a is acceleration]

Put s, u and a in equation :

$\implies$ 1000 = (0 × t) + 1/2 × 5 × t²

$\implies$ 1000 = 0 + 5/2 × t²

$\implies$ 1000 × 2 = 5 × t²

$\implies$ 2000 = 5t²

$\implies$ t² = 2000/5

$\implies$ t² = 400

$\implies$ t = √400

$\implies$ t = 20

Thus,

Time taken by body is 20 seconds.

Answer 2

Information provided with us :

• A body starts from rest and accelerates for 5m/s

What we have to calculate :

• The time taken by the body to cover a distance of 1km

Using Formulas :

Second equation of motion:-

• $\large{\boxed{ \tt{s \: = \: ut + \dfrac{1}{2}at {}^{2} }}}$

Here,

• s is displacement
• a is acceleration
• t is time taken
• u is the initial velocity

Performing Calculations :

• Here we would be substituting the values in the second equation of motion, and we know that intial velocity (u) is always zero . Inorder to calculate the time taken by the body.

Given :

• s = 1km
• a = 5ms-¹
• u = 0

Conversion of unit from km to m :

We know that,

• 1km = 1000 m

So the displacement (s) would be 1000

Putting the values :

$: \longmapsto \: \large{\tt{1000 \: = \: (0)(t) + \dfrac{1}{2}(5)(t {}^{2} )}}$

$: \longmapsto \: \large{\tt{1000 \: = \: (0) \times (t) + \dfrac{1}{2}(5)(t {}^{2} )}}$

$: \longmapsto \: \large{\tt{1000 \: = \: (0) \times (t) + \dfrac{1}{2} \times (5)(t {}^{2} )}}$

$: \longmapsto \: \large{\tt{1000 \: = \: (0) \times (t) + \dfrac{1}{2} \times (5) \times (t {}^{2} )}}$

$: \longmapsto \: \large{\tt{1000 \: = \: \dfrac{1}{2} \times (5) \times (t {}^{2} )}}$

$: \longmapsto \: \large{\tt{1000 \: = \: \dfrac{1}{2} \times 5 \times (t {}^{2} )}}$

$: \longmapsto \: \large{\tt{1000 \: = \: \dfrac{1}{2} \times 5 \times t {}^{2} }}$

$: \longmapsto \: \large{\tt{1000 \: = \: \dfrac{1 \times 5}{2} \times t {}^{2} }}$

$: \longmapsto \: \large{\tt{1000 \: = \: \dfrac{5}{2} \times t {}^{2} }}$

$: \longmapsto \: \large{\tt{1000 \: \times 2 = \:5\times t {}^{2} }}$

$: \longmapsto \: \large{\tt{2000 \: = \:5\times t {}^{2} }}$

$: \longmapsto \: \large{\tt{t {}^{2} \: = \: \dfrac{2000}{5} }}$

$: \longmapsto \: \large{\tt{t {}^{2} \: = \: \cancel\dfrac{2000}{5} }}$

$: \longmapsto \: \large{\tt{t {}^{2} \: = \: 400}}$

$: \longmapsto \: \large{\tt{t {} \: = \: \sqrt{400}} }$

$: \longmapsto \: \boxed{\red{\large{\bf{t {} \: = \: 20}}}}$

$\underline{ \bf{Henceforth, \: time \: taken \: by \: the \: body \: is \: 20s \: to \: cover \: 1000m}}$