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A car starts from rest, accelerates unif...

A car starts from rest, accelerates uniformly for 4 seconds and then moves with uniform velocity . Which of the `(x-t)` graphs shown in fig. represents the motion of the car upto `t=7`s ?

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Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the motion of the car in two distinct phases: the acceleration phase and the uniform velocity phase. ### Step-by-Step Solution: 1. **Understanding the Motion Phases**: - The car starts from rest, which means its initial velocity \( u = 0 \, \text{m/s} \). - It accelerates uniformly for \( t = 4 \, \text{s} \). - After \( 4 \, \text{s} \), the car moves with a constant (uniform) velocity until \( t = 7 \, \text{s} \). 2. **Acceleration Phase (0 to 4 seconds)**: - Since the car is accelerating uniformly, its velocity increases linearly over time. - The final velocity \( v \) at \( t = 4 \, \text{s} \) can be expressed as: \[ v = u + at \] where \( a \) is the acceleration. Since we don't know \( a \) yet, we can denote it as \( a \). 3. **Displacement during Acceleration**: - The displacement \( s \) during the first 4 seconds can be calculated using the equation: \[ s = ut + \frac{1}{2} a t^2 \] Substituting \( u = 0 \): \[ s = \frac{1}{2} a (4^2) = 8a \] 4. **Uniform Velocity Phase (4 to 7 seconds)**: - After 4 seconds, the car moves with a constant velocity \( v \) for \( 3 \, \text{s} \) (from \( t = 4 \, \text{s} \) to \( t = 7 \, \text{s} \)). - The displacement during this phase is: \[ s = v \cdot t = v \cdot 3 = 3v \] - Since \( v = 4a \) (from the previous calculation), the displacement during this phase becomes: \[ s = 3(4a) = 12a \] 5. **Total Displacement**: - The total displacement from \( t = 0 \) to \( t = 7 \) seconds is: \[ \text{Total Displacement} = 8a + 12a = 20a \] 6. **Graphical Representation**: - During the first 4 seconds, the graph of displacement vs. time \( (x-t) \) will be a curve that is increasing in slope (since the velocity is increasing). - From \( t = 4 \, \text{s} \) to \( t = 7 \, \text{s} \), the graph will be a straight line with a constant slope (since the car moves with uniform velocity). 7. **Identifying the Correct Graph**: - We need to find a graph that shows an increasing slope from \( t = 0 \) to \( t = 4 \) seconds and a straight line from \( t = 4 \) to \( t = 7 \) seconds. - Based on the analysis, the correct graph is the one that meets these criteria. ### Conclusion: After evaluating the options, we find that **Option 4** is the correct representation of the car's motion up to \( t = 7 \, \text{s} \).
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