Question

42. A bus moving along a straight line at 20 m/s undergoes an acceleration of 4 m/s2. After 2 seconds, its speed will be :
(a) 8 m/s
(6) 12 m/s
(c) 16 m/s
(d) 28 m/s​

Answer 1

Answer:

(d) 28 m/s

Explanation:

Given :

• Initial velocity = u = 20 m/s

• Acceleration of the bus = a = 4 m/s²

• Time taken = 2 seconds

To find :

• Final velocity of the bus after 2 seconds

Using the first equation of motion :

V=u+at

V=20+4×2

V=20+8

V=28 m/s

The final velocity of the bus is equal to 28 m/s

The correct option to this question is (d)

Answer 2

Given :-

A bus moving along a straight line at 20 m/s undergoes an acceleration of 4 m/s² .

Required to find :-

• Speed of the bus after 2 seconds ?

Concept used :-

• Velocity is a vector quantity

• Velocity needs to have both magnitude and direction

• S.I unit - m/s

• Equations of motion

Equation used :-

$\huge{\dagger{\boxed{\tt{ v = u + at }}}}$

Solution :-

Given that :-

A bus moving along a straight line at 20 m/s undergoes an acceleration of 4 m/s²

We need to find the speed of the bus after 2 seconds !

From the above we can conclude that ;

Initial velocity of the bus = 20 m/s

Acceleration = 4 m/s²

Time = 2 seconds

Using the 1st equation of motion ;

That is

$\huge{\dagger{\boxed{\tt{ v = u + at }}}}$

Here,

v = final velocity

u = initial velocity

a = acceleration

t = time

So,

Substitute the values ;

$\rightarrowtail{\rm{v = 20 \; m/s + ( 4 \; m/s^2 )( 2 \; s )}}$

$\rightarrowtail{\rm{v = 20 \; m/s + 8 \; m/s }}$

$\rightarrowtail{\rm{v = 28 \; m/s }}$

Therefore,

Speed of the bus after 2 seconds = 28m/s

Additional information :-

• The three equations of motion are ;

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

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

$\boxed{\sf{ {v}^{2} - {u}^{2} = 2as }}$

• These are very applicable while solving these type of numericals !