Question

11. A body moves from rest with a uniform acceleration
and travels 270 m in 3 s. Find the velocity of the
body at 10 s after the start.
Ans. 600 m s-1​

Answer 1

Given:

✰ Initial velocity ( u ) = 0 m/s

✰ Distance travelled ( S ) = 270 m

✰ Time taken to travel the given distance ( t ) = 3 s

To find:

✠ The velocity of the body at 10 s after the start.

Solution:

Let a be the uniform acceleration of the body.

Using second equation of motion,

✭ S = ut + 1/2 at²

Putting the values,

⤳ 270 = 0 × 3 + 1/2 × a × 3²

⤳ 270 = 0 + 1/2 × a × 9

⤳ 270 = 0 + 9/2 × a

⤳ 270 = (0 +9)/2 × a

⤳ 270 = 9a/2

⤳ 9a = 270 × 2

⤳ 9a = 540

⤳ a = 540/9

⤳ a = 60 m/s²

Let v be the velocity of the body at 10 s after the start.

Using first equation of motion,

✭ v = u + at

Putting the values,

⤳ v = 0 + 60 × 10

⤳ v = 0 + 600

⤳ v = 600 m/s

∴ 600 m/s is the velocity of the body at 10 s after the start.

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Answer 2

★ The velocity of the body at 10 s after the start is 600 m/s.

Step-by-step explanation

Analysis -

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎It is given that a body moves from rest with a uniform acceleration and travels 270 metres in 3 seconds. So,

• The initial velocity = 0 m/s
• Displacement = 270 m
• Time taken = 3 s

We have been asked to find the velocity of the body after 10 seconds after the start.

Solution -

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Let us assume that the acceleration be 'a'.

And now, we shall apply the second equation of motion to find the uniform acceleration. You might have a small confusion why are we going to use the second equation of motion, not the the first or third. To understand it clearly, have a look at the three equations of motion, mentioned below!

$\: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\boxed{\begin{array}{c} \sf \pmb{1) \: v = u + at \: \: \: \: \: \: \: } \\ \\ \sf \pmb{2) \: s = ut + \dfrac{1}{2}a {t}^{2} } \\ \\ \sf \pmb{3) \: {v}^{2} - {u}^{2} = 2as \: }\end{array}}\end{gathered}$

In the first equation, there are two unknown values, i.e, the acceleration (a) and the final velocity (v) and in the third equation also, we don't know the values of 'v' and 'a'. But in the second equation, only the value of acceleration is unknown. Hence, second equation of motion has to be used.

Now, on substituting the given values in the equation:

$\dashrightarrow{ \sf{s = ut + \dfrac{1}{2} a {t}^{2} }} \\ \\ \\ \dashrightarrow{ \sf{270 = 0 \times 3 + \dfrac{1}{2} \times a \times {3}^{2} }} \\ \\ \\ \dashrightarrow{ \sf{270 = 0 + \dfrac{1}{2} \times a \times 9}} \\ \\ \\ \dashrightarrow{ \sf{270 = \dfrac{1}{2} \times 9 \times a }} \\ \\ \\ \dashrightarrow{ \sf{270 \times 2 = 1 \times 9 \times a}} \\ \\ \\ \dashrightarrow{ \sf{540 = 9 \times a}} \\ \\ \\ \dashrightarrow{ \sf{ \dfrac{ \cancel{540}}{ \cancel9} = a}} \\ \\ \\ \dashrightarrow{ \sf {\pmb{60 \: m \: {s}^{ - 1}} = a}}$

The body was moving 270 m with an acceleration of 60 m/s in 3s.

Since we have been asked to find the velocity of body after 10 s of start, the value of final velocity is required. Now, we will be using the first equation of motion, as it is the most appropriate.

On substituting the values in the equation:

$\dashrightarrow{ \sf{v = u + at}} \\ \\ \\ \dashrightarrow{ \sf{v = 0 + 60 \times 10}} \\ \\ \\ \dashrightarrow{ \sf{v = 0 + 600}} \\ \\ \\ \dashrightarrow \underline{ \boxed{ \sf{ \red {\pmb{v = \frak{600} \: m \: {s}^{ - 1} }}} }}$

And this is the required answer!

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