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An auto travelling along a straight road increases its speed from 30.0 m s-1 to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?
- a)6.0 S
- b)4.5
- c)3.6 s
- d)7.03
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An auto travelling along a straight road increases its speed from 30.0...
Let a be constant acceleration of auto.
Here, u = 30ms−1, v = 50ms−1, S = 180 m
As v2 − u2 = 2aS
Here, u = 30ms−1, v = 50ms−1, S = 180 m
As v2 − u2 = 2aS
Solving this quadratic equation by quadratic formula, we get = 4.5s, −18 s, (t can't be negative)
∴ t = 4.5 s
Most Upvoted Answer
An auto travelling along a straight road increases its speed from 30.0...
To find the time elapsed while the auto moves a distance of 180 m, we can use the equation of motion:
\[v^2 = u^2 + 2as\]
Where:
- \(v\) is the final velocity (50.0 m/s)
- \(u\) is the initial velocity (30.0 m/s)
- \(a\) is the acceleration
- \(s\) is the distance (180 m)
Given that the acceleration is constant, we can rearrange the equation to solve for time:
\[t = \frac{{v - u}}{a}\]
Substituting the given values:
\[t = \frac{{50.0 - 30.0}}{a}\]
Now, we need to find the acceleration. We can use the equation:
\[a = \frac{{v - u}}{t}\]
Let's solve for acceleration first.
Acceleration:
\[a = \frac{{50.0 - 30.0}}{t}\]
\[a = \frac{{20.0}}{t}\]
Substituting this value back into the equation for time:
\[t = \frac{{50.0 - 30.0}}{\frac{{20.0}}{t}}\]
\[t = \frac{{20.0t}}{20.0}\]
\[t^2 = 1\]
\[t = 1\]
Therefore, the time elapsed while the auto moves a distance of 180 m is 1 second.
However, none of the given options match this answer. Let's recheck our calculations.
\[t = \frac{{50.0 - 30.0}}{a}\]
\[t = \frac{{20.0}}{a}\]
Substituting this value back into the equation for distance:
\[s = ut + \frac{1}{2}at^2\]
\[180 = 30t + \frac{1}{2}at^2\]
\[180 = 30 \left(\frac{{20.0}}{a}\right) + \frac{1}{2}a \left(\frac{{20.0}}{a}\right)^2\]
\[180 = \frac{{600}}{a} + \frac{{200}}{a}\]
\[180 = \frac{{800}}{a}\]
\[a = \frac{{800}}{180}\]
\[a = 4.44 \, \text{m/s}^2\]
Now, let's calculate the time again:
\[t = \frac{{20.0}}{a}\]
\[t = \frac{{20.0}}{4.44}\]
\[t \approx 4.50\]
Therefore, the correct answer is option B) 4.5 seconds.
\[v^2 = u^2 + 2as\]
Where:
- \(v\) is the final velocity (50.0 m/s)
- \(u\) is the initial velocity (30.0 m/s)
- \(a\) is the acceleration
- \(s\) is the distance (180 m)
Given that the acceleration is constant, we can rearrange the equation to solve for time:
\[t = \frac{{v - u}}{a}\]
Substituting the given values:
\[t = \frac{{50.0 - 30.0}}{a}\]
Now, we need to find the acceleration. We can use the equation:
\[a = \frac{{v - u}}{t}\]
Let's solve for acceleration first.
Acceleration:
\[a = \frac{{50.0 - 30.0}}{t}\]
\[a = \frac{{20.0}}{t}\]
Substituting this value back into the equation for time:
\[t = \frac{{50.0 - 30.0}}{\frac{{20.0}}{t}}\]
\[t = \frac{{20.0t}}{20.0}\]
\[t^2 = 1\]
\[t = 1\]
Therefore, the time elapsed while the auto moves a distance of 180 m is 1 second.
However, none of the given options match this answer. Let's recheck our calculations.
\[t = \frac{{50.0 - 30.0}}{a}\]
\[t = \frac{{20.0}}{a}\]
Substituting this value back into the equation for distance:
\[s = ut + \frac{1}{2}at^2\]
\[180 = 30t + \frac{1}{2}at^2\]
\[180 = 30 \left(\frac{{20.0}}{a}\right) + \frac{1}{2}a \left(\frac{{20.0}}{a}\right)^2\]
\[180 = \frac{{600}}{a} + \frac{{200}}{a}\]
\[180 = \frac{{800}}{a}\]
\[a = \frac{{800}}{180}\]
\[a = 4.44 \, \text{m/s}^2\]
Now, let's calculate the time again:
\[t = \frac{{20.0}}{a}\]
\[t = \frac{{20.0}}{4.44}\]
\[t \approx 4.50\]
Therefore, the correct answer is option B) 4.5 seconds.
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An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer?
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An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer?.
An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An auto travelling along a straight road increases its speed from 30.0 m s-1to 50.0 m s-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?a)6.0 Sb)4.5c)3.6 sd)7.03Correct answer is option 'B'. Can you explain this answer?.
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