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A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .?
Most Upvoted Answer
A car starts from rest and moves with a uniform acceleration of 1 m/s^...
Given information:
- Initial velocity (u) = 0 m/s (car starts from rest)
- Acceleration (a) = 1 m/s^2 (uniform acceleration for 10 seconds)
- Time (t1) = 10 seconds (acceleration time)
- Time (t2) = 30 seconds (uniform velocity time)
- Time (t3) = 10 seconds (braking time)
Calculating distance during acceleration:
Using the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
After 10 seconds of acceleration, the final velocity can be calculated as:
v = u + at
v = 0 + 1 * 10
v = 10 m/s
The distance covered during acceleration can be calculated using the formula:
s = ut + (1/2)at^2
where s is the distance covered.
Substituting the values, we get:
s1 = 0 * 10 + (1/2) * 1 * (10)^2
s1 = 0 + (1/2) * 1 * 100
s1 = 50 meters
Calculating distance during uniform velocity:
During the 30 seconds of uniform velocity, the car maintains a constant speed of 10 m/s. The distance covered can be calculated as the product of the velocity and time:
s2 = 10 * 30
s2 = 300 meters
Calculating distance during braking:
When the brakes are applied, the car decelerates uniformly until it comes to rest. The final velocity during braking is 0 m/s. Using the equation of motion:
v = u + at
Substituting the values, we get:
0 = 10 + (-1) * 10
0 = 10 - 10
0 = 0
The negative sign indicates deceleration. The distance covered during braking can be calculated using the formula:
s = ut + (1/2)at^2
where s is the distance covered.
Substituting the values, we get:
s3 = 10 * 10 + (1/2) * (-1) * (10)^2
s3 = 100 - 50
s3 = 50 meters
Total distance covered:
The total distance covered by the car is the sum of the distances during acceleration, uniform velocity, and braking:
Total distance = s1 + s2 + s3
Total distance = 50 + 300 + 50
Total distance = 400 meters
Therefore, the total distance covered by the car is 400 meters.
- Initial velocity (u) = 0 m/s (car starts from rest)
- Acceleration (a) = 1 m/s^2 (uniform acceleration for 10 seconds)
- Time (t1) = 10 seconds (acceleration time)
- Time (t2) = 30 seconds (uniform velocity time)
- Time (t3) = 10 seconds (braking time)
Calculating distance during acceleration:
Using the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
After 10 seconds of acceleration, the final velocity can be calculated as:
v = u + at
v = 0 + 1 * 10
v = 10 m/s
The distance covered during acceleration can be calculated using the formula:
s = ut + (1/2)at^2
where s is the distance covered.
Substituting the values, we get:
s1 = 0 * 10 + (1/2) * 1 * (10)^2
s1 = 0 + (1/2) * 1 * 100
s1 = 50 meters
Calculating distance during uniform velocity:
During the 30 seconds of uniform velocity, the car maintains a constant speed of 10 m/s. The distance covered can be calculated as the product of the velocity and time:
s2 = 10 * 30
s2 = 300 meters
Calculating distance during braking:
When the brakes are applied, the car decelerates uniformly until it comes to rest. The final velocity during braking is 0 m/s. Using the equation of motion:
v = u + at
Substituting the values, we get:
0 = 10 + (-1) * 10
0 = 10 - 10
0 = 0
The negative sign indicates deceleration. The distance covered during braking can be calculated using the formula:
s = ut + (1/2)at^2
where s is the distance covered.
Substituting the values, we get:
s3 = 10 * 10 + (1/2) * (-1) * (10)^2
s3 = 100 - 50
s3 = 50 meters
Total distance covered:
The total distance covered by the car is the sum of the distances during acceleration, uniform velocity, and braking:
Total distance = s1 + s2 + s3
Total distance = 50 + 300 + 50
Total distance = 400 meters
Therefore, the total distance covered by the car is 400 meters.
Community Answer
A car starts from rest and moves with a uniform acceleration of 1 m/s^...
Here,
t1= 10 s
u=0. (rest )
a=1m/s
therefore,
S=ut+1/2at^2
=(0)10+1/2(1)(100^2)
=0+5000
=5000 (1)
NOW
using 3rd equation of motion we will find out V
v^2 = u^2+2as
v^2=0^2+2(1)(5000)
=10,000m/s
With this velocity it continues for 30 seconds . Therefore with t2 = 30s
we will find out S
S= ut+1/2at^2
=10,000(30)+1/2(1)(30^2)
=150,450 (2)
NOW breaks are applied therefore there is retardation i.e. a will be negitive
S=ut+1/2at^2
=10,000(10)+1/2(-1)(10^2)
=100,000+(-50)
=100,000-50
=99,950 (3)
WE have (1),(2),(3) , add all the distance
5000+150,450+99,50=
255,400
t1= 10 s
u=0. (rest )
a=1m/s
therefore,
S=ut+1/2at^2
=(0)10+1/2(1)(100^2)
=0+5000
=5000 (1)
NOW
using 3rd equation of motion we will find out V
v^2 = u^2+2as
v^2=0^2+2(1)(5000)
=10,000m/s
With this velocity it continues for 30 seconds . Therefore with t2 = 30s
we will find out S
S= ut+1/2at^2
=10,000(30)+1/2(1)(30^2)
=150,450 (2)
NOW breaks are applied therefore there is retardation i.e. a will be negitive
S=ut+1/2at^2
=10,000(10)+1/2(-1)(10^2)
=100,000+(-50)
=100,000-50
=99,950 (3)
WE have (1),(2),(3) , add all the distance
5000+150,450+99,50=
255,400
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A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .?
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A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .?.
A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starts from rest and moves with a uniform acceleration of 1 m/s^2 for 10 seconds . It then moves with a uniform velocity for 30 seconds . The brakes are then applied and the car is brought to rest in 10 seconds . Find the total distance covered by the car .?.
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