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प्रश्न
A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
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उत्तर
The data given in this numerical problem are in different units. So, we should first convert km h-1 unit into m s-1 unit.
For first car:
Initial velocity u = 52 km h-1
= `(52 "km")/(1 "h")`
= `(52 xx 1000 "m")/(1 xx 3600 "s")`
= 14.4 s-1
Final velocity, v = 0 km h-1 = 0.0 m s-1
Time taken, t = 5s
For second car:
Initial velocity, u = 3 km h-1
= `(3 "km")/(1"h")`
= `(3 xx 1000 "m")/(1 xx 3600 "s")`
= 0.8 ms-1
Final velocity, v = 0 km h-1 = 0.0 m s-1
Time taken, t = 10s

The area under a moving body's speed-time graph indicates the distance it has traveled.
So, Distance travelled by the first car = Area of the triangle AOB
= `1/2 xx "OB" xx "AO"`
= `1/2 xx 14.4` ms-1 × 5s
= `1/2 xx 14.4 xx 5 "m"`
= 36 m
Similarly, distance travelled by the second car = area of triangle COD.
= `1/2 xx "OD" xx "CO"`
= `1/2 xx 0.83 m s^-1 xx 10 s`
= `1/2 xx 0.83 xx 10 "m"`
= 4.1 m
Thus, the second car travels 4.1 m and the first car travels 36 m before coming to rest.
So, the first car travelled farther after the brakes were applied.
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संबंधित प्रश्न
The speed-time graph for a car is shown in the following figure:

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Fill in the following blank with suitable word :
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Two students were asked to plot a distance-time graph for the motion described by Table A and Table B.
| Table A |
||||||
| Distance moved (m) | 0 | 10 | 20 | 30 | 40 | 50 |
| Time (minutes) | 0 | 2 | 4 | 6 | 8 | 10 |
| Table B |
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| Distance moved (m) | 0 | 5 | 10 | 15 | 20 | 25 |
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 |

The graph given in figure is true for
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