A body move from rest with a constant acceleration of 5 m/sec2. The distance cover in 5 sec is

This question was previously asked in
ISRO VSSC Technical Assistant Mechanical 8 Feb 2015 Official Paper
View all ISRO Technical Assistant Papers >
  1. 12.5 m
  2. 62.5 m
  3. 125 m
  4. 130 m

Answer (Detailed Solution Below)

Option 2 : 62.5 m
Free
ISRO Technical Assistant Mechanical Full Mock Test
2.7 K Users
80 Questions 80 Marks 90 Mins

Detailed Solution

Download Solution PDF

Concept:

Equation of motion:

It is defined as equations that describe the behaviour of a physical system in terms of its motion as a function of time.

There are three equation of motions:

  1. v = u + at
  2. \(s = ut + \frac 12at^2\)
  3. v2 = u2 + 2as

where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time

Calculation:

Given:

u = 0 m/s, a = 5 m/s2, t = 5 s

\(s = ut + \frac 12at^2\)

\(s = 0\times t + \frac 12at^2\)

\(s = \frac 12at^2 = \frac 12 \times 5\times 5^2= 62.5 ~m\)

Latest ISRO Technical Assistant Updates

Last updated on May 30, 2025

-> The ISRO Technical Assistant recruitment 2025 notification has been released at the official website. 

-> Candidates can apply for ISRO recruitment 2025 for Technical Assistant from June 4 to 18.

-> A total of 83 vacancies have been announced for the post of Technical Assistant.

-> The Selection process consists of a written test and a Skill test.

-> Candidates can also practice through ISRO Technical Assistant Electrical Test Series, ISRO Technical Assistant Electronics Test Series, and ISRO Technical Assistant Mechanical Test Series to improve their preparation and increase the chance of selection. 

More Force Mass and Acceleration Questions

More Kinematics and Kinetics Questions

Get Free Access Now
Hot Links: teen patti rummy 51 bonus teen patti 3a teen patti glory