
A body moving with uniform acceleration \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\] starts from rest. The distance covered by it in fifth second will be
A. \[8{\rm{ m}}\]
B. \[64{\rm{ m}}\]
C. \[4{\rm{ m}}\]
D. \[36{\rm{ m}}\]
Answer
589.5k+ views
Hint: Distance travelled by a body in the nth second will be calculated in this solution which gives us the relation between the number of the second in which distance is to be calculated, initial velocity and the acceleration.
Complete step by step answer:
The initial velocity of the body is u.
The final velocity of the body is v.
The acceleration of the body is \[a = 8{\rm{ m}}{{\rm{s}}^{ - 2}}\].
It is given that the body is moving with uniform acceleration \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\] and it starts from the rest so the initial velocity of the given body is zero.
It is known to us that the expression for displacement in the nth second is given by the relation:
\[s = u + a\left( {n - \dfrac{1}{2}} \right)\]……(1)
Here n is the number of seconds for which the distance travelled by the body is to be evaluated.
Distance covered by the body in fifth second moving with uniform acceleration is to be calculated.
Substitute \[5\] for n, \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\] for a and \[0\] for u in equation (1).
\[\begin{array}{l}
s = 0 + 8{\rm{ m}}{{\rm{s}}^{ - 2}}\left( {5 - \dfrac{1}{2}} \right)\\
= 36{\rm{ m}}
\end{array}\]
Here s gives us the value of the distance covered by the body in the fifth second with a uniform acceleration \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\].
Therefore, the distance covered in fifth second is 36 m
So, the correct answer is “Option D”.
Note:
Do not use the second equation of motion in the solution of this problem because in that equation time period t is to be substituted but here distance is to be calculated for fifth second.
Complete step by step answer:
The initial velocity of the body is u.
The final velocity of the body is v.
The acceleration of the body is \[a = 8{\rm{ m}}{{\rm{s}}^{ - 2}}\].
It is given that the body is moving with uniform acceleration \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\] and it starts from the rest so the initial velocity of the given body is zero.
It is known to us that the expression for displacement in the nth second is given by the relation:
\[s = u + a\left( {n - \dfrac{1}{2}} \right)\]……(1)
Here n is the number of seconds for which the distance travelled by the body is to be evaluated.
Distance covered by the body in fifth second moving with uniform acceleration is to be calculated.
Substitute \[5\] for n, \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\] for a and \[0\] for u in equation (1).
\[\begin{array}{l}
s = 0 + 8{\rm{ m}}{{\rm{s}}^{ - 2}}\left( {5 - \dfrac{1}{2}} \right)\\
= 36{\rm{ m}}
\end{array}\]
Here s gives us the value of the distance covered by the body in the fifth second with a uniform acceleration \[8{\rm{ m}}{{\rm{s}}^{ - 2}}\].
Therefore, the distance covered in fifth second is 36 m
So, the correct answer is “Option D”.
Note:
Do not use the second equation of motion in the solution of this problem because in that equation time period t is to be substituted but here distance is to be calculated for fifth second.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

