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A car starts from rest and move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?
Verified Answer
A car starts from rest and move along x-axis with constant acceleratio...
V=u+at
v=0+5m/s^2*8
v=40m/s
constant velocity=40m/s
distance covered in the first 8 sec = S=1/2a*t^2=160m.
distance covered in the last 4 sec=const. speed*time=40m/s*4=160m
total distance covered in 12 sec=160+160=320m ans.
ans=320m
This question is part of UPSC exam. View all Class 9 courses
Most Upvoted Answer
A car starts from rest and move along x-axis with constant acceleratio...
Calculation of Distance Covered
To find the distance covered by the car in 12 seconds since starting from rest, we need to break down the motion into two parts: the first 8 seconds with constant acceleration and the remaining 4 seconds with constant velocity.
1. Motion with Constant Acceleration
Given:
Initial velocity, u = 0 m/s
Acceleration, a = 5 m/s²
Time, t₁ = 8 s
Using the equation of motion:
s = ut + (1/2)at²
Substituting the given values:
s₁ = 0(8) + (1/2)(5)(8)²
s₁ = 0 + (1/2)(5)(64)
s₁ = 0 + (1/2)(320)
s₁ = 0 + 160
s₁ = 160 m
Therefore, the distance covered during the first 8 seconds with constant acceleration is 160 meters.
2. Motion with Constant Velocity
After 8 seconds, the car continues to move with constant velocity. Since the car started from rest, its final velocity at the end of 8 seconds would be:
v = u + at
v = 0 + (5)(8)
v = 40 m/s
Now, during the next 4 seconds, the car maintains this constant velocity of 40 m/s. Therefore, the distance covered during this time can be calculated using the formula:
s = vt
Substituting the values:
s₂ = (40)(4)
s₂ = 160 m
Therefore, the distance covered during the next 4 seconds with constant velocity is 160 meters.
Total Distance Covered
To find the total distance covered in 12 seconds, we add the distances covered during the two phases:
Total distance = s₁ + s₂
Total distance = 160 + 160
Total distance = 320 m
Therefore, the car will cover a distance of 320 meters in 12 seconds since it started from rest.
To find the distance covered by the car in 12 seconds since starting from rest, we need to break down the motion into two parts: the first 8 seconds with constant acceleration and the remaining 4 seconds with constant velocity.
1. Motion with Constant Acceleration
Given:
Initial velocity, u = 0 m/s
Acceleration, a = 5 m/s²
Time, t₁ = 8 s
Using the equation of motion:
s = ut + (1/2)at²
Substituting the given values:
s₁ = 0(8) + (1/2)(5)(8)²
s₁ = 0 + (1/2)(5)(64)
s₁ = 0 + (1/2)(320)
s₁ = 0 + 160
s₁ = 160 m
Therefore, the distance covered during the first 8 seconds with constant acceleration is 160 meters.
2. Motion with Constant Velocity
After 8 seconds, the car continues to move with constant velocity. Since the car started from rest, its final velocity at the end of 8 seconds would be:
v = u + at
v = 0 + (5)(8)
v = 40 m/s
Now, during the next 4 seconds, the car maintains this constant velocity of 40 m/s. Therefore, the distance covered during this time can be calculated using the formula:
s = vt
Substituting the values:
s₂ = (40)(4)
s₂ = 160 m
Therefore, the distance covered during the next 4 seconds with constant velocity is 160 meters.
Total Distance Covered
To find the total distance covered in 12 seconds, we add the distances covered during the two phases:
Total distance = s₁ + s₂
Total distance = 160 + 160
Total distance = 320 m
Therefore, the car will cover a distance of 320 meters in 12 seconds since it started from rest.
Community Answer
A car starts from rest and move along x-axis with constant acceleratio...
Wow cool...ur answer was just awesome
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A car starts from rest and move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?
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A car starts from rest and move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest? 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 move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest? 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 move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?.
A car starts from rest and move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest? 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 move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest? 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 move along x-axis with constant acceleration 5m per second square for 8 seconds. if it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?.
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