Question 5 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?
Class 9 - Science - Motion Page 112
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Answered by
5
Given for first driver,
Intital velocity, u=52 kmh−1u=52 kmh-1
=52×1000m60×60s=14.4 m s−1=52×1000m60×60s=14.4 m s-1
Time, t=5st=5s
Final velocity v=0v=0 (Since car stops)
Therefore, distance, s=s= ?
Given for second driver, u=3 km h−1u=3 km h-1
=3000m60×60s=9.4 ms−1=3000m60×60s=9.4 ms-1 `
Time, t=10st=10s
Final velocity v=0v=0
In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.
Distance is calculated by the area under the slope of the graph.
Thus, distance covered by 1st car = Area of △OAD△OAD
⇒ Distance s=12×OD×OAs=12×OD×OA
⇒s=12×14.4 m/s×5s⇒s=12×14.4 m/s×5s
⇒s=7.2 m/s×5s=36m⇒s=7.2 m/s×5s=36m
Thus, distance covered by 2nd car = Area of △OBC
⇒ Distance, s=12×OC×OBs=12×OC×OB
⇒s=12×9.4 m/s×10s⇒s=12×9.4 m/s×10s
⇒s=4.7 m/s×10s=47m⇒s=4.7 m/s×10s=47m
Therefore, second car travelled farther
Intital velocity, u=52 kmh−1u=52 kmh-1
=52×1000m60×60s=14.4 m s−1=52×1000m60×60s=14.4 m s-1
Time, t=5st=5s
Final velocity v=0v=0 (Since car stops)
Therefore, distance, s=s= ?
Given for second driver, u=3 km h−1u=3 km h-1
=3000m60×60s=9.4 ms−1=3000m60×60s=9.4 ms-1 `
Time, t=10st=10s
Final velocity v=0v=0
In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.
Distance is calculated by the area under the slope of the graph.
Thus, distance covered by 1st car = Area of △OAD△OAD
⇒ Distance s=12×OD×OAs=12×OD×OA
⇒s=12×14.4 m/s×5s⇒s=12×14.4 m/s×5s
⇒s=7.2 m/s×5s=36m⇒s=7.2 m/s×5s=36m
Thus, distance covered by 2nd car = Area of △OBC
⇒ Distance, s=12×OC×OBs=12×OC×OB
⇒s=12×9.4 m/s×10s⇒s=12×9.4 m/s×10s
⇒s=4.7 m/s×10s=47m⇒s=4.7 m/s×10s=47m
Therefore, second car travelled farther
Answered by
14
→ For First Driver, Initial velocity (u) = 52 Km h-¹
= 52 × 1000m/ 60 × 60s
= 14.4 ms-¹
Final velocity of first diver (v) = 0
Time Taken , (t) = 5 s.
Therefore, Distance (s) = ?
→ For Second Diver, Initial velocity (u) = 3 Km h-¹
= 3000m/ 60 × 60s
= 9.4 ms-¹
Final velocity of second driver , (v) = 0
Time Taken, (t) = 10 s.
In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.
Distance is calculated by the area under the slope of the graph.
Therefore, Distance travelled by 1st Car = Area of ∆ OAD
⇒ Distance (s) = 1/2 × OD × OA
⇒ s = 1/2 × 14.4 m/s × 5s
⇒s = 7.2 m/s × 5s
= 36 m.
Similarly, Distance travelled by 2nd car = Area of ∆ OBC
⇒ Distance (s) = 1/2 × OC × OD
⇒ s = 1/2 × 9.4m/s × 10s
⇒ s = 4.7 m/s × 10s
= 47 m.
Therefore, The second Car travelled more Distance than First car ,after the brakes were applied.
= 52 × 1000m/ 60 × 60s
= 14.4 ms-¹
Final velocity of first diver (v) = 0
Time Taken , (t) = 5 s.
Therefore, Distance (s) = ?
→ For Second Diver, Initial velocity (u) = 3 Km h-¹
= 3000m/ 60 × 60s
= 9.4 ms-¹
Final velocity of second driver , (v) = 0
Time Taken, (t) = 10 s.
In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.
Distance is calculated by the area under the slope of the graph.
Therefore, Distance travelled by 1st Car = Area of ∆ OAD
⇒ Distance (s) = 1/2 × OD × OA
⇒ s = 1/2 × 14.4 m/s × 5s
⇒s = 7.2 m/s × 5s
= 36 m.
Similarly, Distance travelled by 2nd car = Area of ∆ OBC
⇒ Distance (s) = 1/2 × OC × OD
⇒ s = 1/2 × 9.4m/s × 10s
⇒ s = 4.7 m/s × 10s
= 47 m.
Therefore, The second Car travelled more Distance than First car ,after the brakes were applied.
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