Question

An object starts from rest with uniform acceleration and covers 100 m in 10s.Find the speed of the object just after covering 50 m.​

Answer 1

Given:

Initial velocity = 0 m/s

Distance travelled in 10 sec = 500 m

To find:

Speed of object after covering 50 m ?

Concept:

Since the acceleration is assigned to be constant in the question , we can easily apply Equations of Kinematics to solve this question.

Calculation:

$s = ut + \dfrac{1}{2} a {t}^{2}$

$= > 100 = (0 \times 10) + \dfrac{1}{2} \times a \times {(10)}^{2}$

$= > 100 = \dfrac{1}{2} \times a \times {(10)}^{2}$

$= > \cancel{100} = \dfrac{1}{2} \times a \times \cancel{100}$

$= > a = 2 \: m {s}^{ - 2}$

Now for 50 m distance journey :

Let final Velocity be v :

${v}^{2} = {u}^{2} + 2as$

$= > {v}^{2} = {(0)}^{2} +( 2 \times 2 \times 50)$

$= > {v}^{2} = 200$

$= > v = 10 \sqrt{2} \: m {s}^{ - 1}$

So final answer is:

Final Velocity after 50 m = 10√2 m/s

Answer 2

AnswEr :

Given that an object starting from rest with uniform acceleration , covers 100m in 10s.

We have to find speed of object after covering 50m.

From the given data : Initial speed (u) = 0 m/s. Distance (s) = 100m. Time (t) = 10s.

At first to find acceleration we will use 2nd equation of motion i.e. s = ut + ½ at²

Substituting values,

→ 100 = 0 × 10 + ½ × a × 10²

→ 100 = 0 + ½ × a × 100

→ 100 = 50a

→ a = 100/50

→ a = 2 m/s²

Now we have acceleration i.e. a = 2 m/s². Now to find final velocity we can use 3rd equation of motion as distance has been given particularly i.e. v² - u² = 2as

Substituting values,

→ v² - 0² = 2 × 2 × 50

→ v² - 0 = 200

→ v² = 200

→ v = √200

→ v = 14.14 m/s

∴ Final speed of object = 14.14 m/s (Ans.)