JEE Exam  >  JEE Questions  >  A car starts moving rectilinearly, first with... Start Learning for Free
A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly?
Verified Answer
A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s...
Ans.


This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s...
Analysis:
We are given the following information:
- Initial acceleration (ω) = 5.0 m/s^2
- Initial velocity (v0) = 0 m/s
- Total time of motion (τ) = 25 s
- Average velocity (v) = 72 km/h

We need to determine the duration for which the car moves uniformly.

Step 1: Convert average velocity to m/s
Given that the average velocity is 72 km/h, we need to convert it to m/s to match the units used for acceleration and time.
1 km/h = (1000 m / 3600 s) = 0.2778 m/s
Therefore, the average velocity (v) = 72 km/h * 0.2778 m/s = 20 m/s.

Step 2: Determine the distance traveled
Using the formula for average velocity, we can calculate the distance traveled by the car.
Average velocity (v) = total distance / total time
Total distance = average velocity * total time
Total distance = 20 m/s * 25 s = 500 m

Step 3: Determine the time taken for acceleration and deceleration
During the acceleration and deceleration phases, the velocity changes from 0 m/s to some final velocity and then back to 0 m/s. The total distance covered during these phases can be determined using the equations of motion.

For acceleration:
Final velocity (v) = initial velocity (v0) + acceleration (ω) * time
0 m/s = 0 m/s + 5.0 m/s^2 * t1
t1 = 0 s (since the car starts from rest)

For deceleration:
Final velocity (v) = initial velocity (v0) + acceleration (ω) * time
0 m/s = v1 + 5.0 m/s^2 * t2
v1 = -5.0 m/s * t2

The total distance covered during acceleration and deceleration can be calculated using the equation:
Total distance = (initial velocity + final velocity) * time / 2

Total distance = (0 m/s + v1) * t2 / 2 + (0 m/s + 0 m/s) * t1 / 2
Total distance = v1 * t2 / 2

Substituting the value of v1 from the deceleration equation:
Total distance = (-5.0 m/s * t2) * t2 / 2
Total distance = -2.5 m/s * t2^2

Step 4: Determine the time taken for uniform motion
The remaining distance covered by the car is during the uniform motion phase.
Total distance = distance during acceleration + distance during uniform motion + distance during deceleration
500 m = 0 m + distance during uniform motion + (-2.5 m/s * t2^2)

Simplifying the equation:
500 m = -2.5 m/s * t2^2

Key Point:
Since time cannot be negative, the negative sign is ignored for simplification purposes.

Step 5: Solve for time
Rearranging the equation:
t2^2 = 500 m / 2.5 m
Explore Courses for JEE exam
A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly?
Question Description
A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly?.
Solutions for A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? defined & explained in the simplest way possible. Besides giving the explanation of A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly?, a detailed solution for A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? has been provided alongside types of A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? theory, EduRev gives you an ample number of questions to practice A car starts moving rectilinearly, first with acceleration ω = 5.0 m/s2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate ω, comes to a stop. The total time of motion equals τ = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev