Question

a cyclist cover 56 and 58 meters in 20th , 21th second respectively. the body starts options a) from rest and move in a uniform velocity option b) from rest and move in a uniform acceleration option c) with an initial velocity and moves with uniform acceleration option d)with an initial velocity and moves with uniform velocity​

Answer 1

A cyclist cover 56 and 58 meters in 20th, 21st second respectively. the body starts  

a) from rest and move in a uniform velocity

b) from rest and move in a uniform acceleration

c) with an initial velocity and moves with uniform acceleration

d)with an initial velocity and moves with uniform velocity​

[Complete question]

Given:

A cyclist cover 56 and 58 meters in 20th, 21st second respectively.

To find:

The correct option for the given condition

Solution:

As the cyclist increases his speed by 2m in one second so this can be the case of the uniform acceleration which is defined as: "The rate of the change of the speed of the body remains constant"

So the cyclist will move with uniform acceleration. Now we will check the cyclist started from rest or he was moving with some initial velocity.

As per the given information cyclist cover distance:

in 20th second is 56 meters distance and in the very next second, it covers 21st second is 58 meters distance.

Let us find the value of the initial velocity to get it clear whether the object was moving or at rest.

By the second law of the motion, we have

$S = ut + (\frac{1}{2} )at^{2}$

For the first case:

T1 = 19 sec

T2 = 20 sec

S = 56 m

Distance travelled in 20th sec  = Distance covered in 20 sec - Distance covered in 19 sec

$56 = 20u + (\frac{1}{2} )a(20)^{2}- (19u +(\frac{1}{2} )a(19)^{2})$

On solving the above equation we get:

$56 = u + (\frac{39}{2} )a$---(I)

For the second case:

T1 = 20 sec

T2 = 21 sec

S = 58 m

Distance covered in 21st sec

=Distance covered in 21 sec - Distance covered in 20 sec

$58 = 21u + (\frac{1}{2} )a(21)^{2}- (20u +(\frac{1}{2} )a(20)^{2})$

$58 = u + (\frac{41}{2} )a$--------(II)

(II) - (I) we get the value of the acceleration as:

58 - 56 = u-u + (41-39)a/2

a = 2

$56 = u + (\frac{39}{2} )a$

56 = u + 39

u  = 17  

so the Initial Velocity = 17 m/sec

So the correct option will be:

c) with an initial velocity of 17 m/sec and moves with uniform acceleration

Answer 2

Given : a cyclist cover 56 and 58 meters in 20th , 21th second respectively

To Find : Choose correct option :

a) from rest and move in a uniform velocity option

b) from rest and move in a uniform acceleration option

c) with an initial velocity and moves with uniform acceleration option d)with an initial velocity and moves with uniform velocity​

Solution:

cyclist cover 56 and 58 meters in 20th , 21th second respectively.

As distance covered in 1 sec is varying hence uniform velocity is not possible so option a & d are not possible

so assuming uniform acceleration option b or c .

S = ut + (1/2)at²

Distance covered in 20th sec

Distance covered in 20 sec - Distance covered in 19 sec

=> 56 = 20u + (1/2)a(20)²  -  (19u + (1/2)a(19)²

=> 56 = u  + (39/2)a    Eq1

Distance covered in 21st sec

Distance covered in 21 sec - Distance covered in 20 sec

=> 58 = 21u + (1/2)a(21)²  -  (20u + (1/2)a(20)²

=> 58 = u  + (41/2)a   Eq2

Eq2 - Eq1

=>  2 =  a

56 = u  + (39/2)a  

=> 56 = u + 39

=> u  = 17  

Initial Velocity = 17 m/s

Correct option :  c) with an initial velocity and moves with uniform acceleration option

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