Question

A bus moving along a straight line at 20m/s under goes an acceleration of 4m/s. After 2 sec ,it's speed will be. (I) 8m/s (ii) 12m/s (III) 16m/s (IV)28m/s

Answer 1

Given

• u = 20 m/s
• a = 4 m/s²

To Find

• Final velocity

Solution

☯ v = u + at

• Here we use the first equation of kinematics

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✭ According to the Question :

➞ v = u + at

• v = ?
• u = 20 m/s
• a = 4 m/s²
• t = 2 sec

➞ v = 20 + 4(2)

➞ v = 20 + 8

➞ v = 28 m/s

∴ The final velocity is 28 m/s

Answer 2

Solution:

Final velocity(v) = ?

Initial velocity(u) = 20m/s

Acceleration(a) = 4m/s²

Time(t) = 2s

Using 1st equation of motion,

$\tt \underline{v = u + at}$

$\leadsto \sf{v =20+4 \times 2}$

$\leadsto \sf{v = 20 + 8}$

$\leadsto \small\pmb{v = 28m/s}$

$\tt\therefore{Final \: velocity = 28m/s}$

Hence, the speed of bus after 2s will be 28m/s.

Thus, option no. (iv) is the correct answer.

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Additional Information:

Equations of motion

Relation between initial velocity, final velocity, acceleration, time:

If initial velocity of a moving body is u

and after t seconds its velocity becomes v with uniform acceleration

a and the distance travelled during this time is s, the equations of motion are as follows:

1. v = u + at
2. s = ut + 1/2 at²
3. v² = u² + 2as

When body is comming down towards the earth, the equation of motion can be written as:

1. v = gt
2. h = ut + 1/2 gt²
3. v² = u² + 2gh

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