Let one side distance = x km. Time taken for forward trip at a speed of 20 km/h = Distance / Speed = x/20 h. Time taken in return trip at a speed of 30 km/h = x/30 h. Total time for the whole trip =𝑥/20+𝑥/30=3𝑥+2𝑥/60=5𝑥/60 h. Total distance covered = 2x km. We know, Average speed = Total distance ÷Read more
Let one side distance = x km.
Time taken for forward trip at a speed of 20 km/h = Distance / Speed = x/20 h.
Time taken in return trip at a speed of 30 km/h = x/30 h.
Total time for the whole trip =𝑥/20+𝑥/30=3𝑥+2𝑥/60=5𝑥/60 h.
Total distance covered = 2x km.
We know, Average speed = Total distance ÷ Total time
= 2x ÷ (5x/60) = 24 kmh⁻¹.
(a) For motion from A to B: Distance covered = 300 m Displacement = 300 m. Time taken = 150 sec. We know that, Average speed = Total distance covered ÷ Total time taken = 300 m ÷ 150 sec = 2 ms⁻¹ Average velocity = Net displacement ÷ time taken = 300 m ÷ 150 sec = 2 ms⁻¹ (b) For motion from A to C:Read more
(a) For motion from A to B:
Distance covered = 300 m
Displacement = 300 m.
Time taken = 150 sec.
We know that, Average speed = Total distance covered ÷ Total time taken
= 300 m ÷ 150 sec = 2 ms⁻¹
Average velocity = Net displacement ÷ time taken
= 300 m ÷ 150 sec = 2 ms⁻¹
(b) For motion from A to C:
Distance covered = 300 + 100 = 400 m.
Displacement = AB – CB = 300 – 100 = 200 m.
Time taken = 2.5 min + 1 min = 3.5 min = 210 sec.
Therefore, Average speed = Total distance covered ÷ Total time taken
= 400 ÷ 210 = 1.90 ms⁻¹.
Average velocity = Net displacement ÷ time taken
= 200 m ÷ 210 sec = 0.952ms⁻¹.
Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. In 40 sec the athlete complete one round. So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round. ⇒ Distance covered in 140 sec = 2πr × 3.5 = 2 × 22/7 × 100 × 3.5 = 2200 m. At the end of his motion, the athlete will be in the diametricRead more
Time taken = 2 min 20 sec = 140 sec.
Radius, r = 100 m.
In 40 sec the athlete complete one round.
So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.
⇒ Distance covered in 140 sec = 2πr × 3.5 = 2 × 22/7 × 100 × 3.5 = 2200 m.
At the end of his motion, the athlete will be in the diametrically opposite position.
⇒ Displacement = diameter = 200 m.
Here we have, Initial velocity (u) = 5 m/s Final velocity (v) = 0 m/s Acceleration (a) = – 10 m/s² Height, i.e. Distance, s =? Time (t) taken to reach the height =? We know that, v² = u² + 2as ⇒ 0 = (5) ² + 2 × −10 × s ⇒ 0 = 25 − 20s ⇒ s = 25/20 m ⇒ s = 1.25 m Now, we know that, v = u + at ⇒ 0 = 5 +Read more
Here we have,
Initial velocity (u) = 5 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = – 10 m/s²
Height, i.e. Distance, s =?
Time (t) taken to reach the height =?
We know that, v² = u² + 2as
⇒ 0 = (5) ² + 2 × −10 × s
⇒ 0 = 25 − 20s
⇒ s = 25/20 m
⇒ s = 1.25 m
Now, we know that, v = u + at
⇒ 0 = 5 + (–10) × t
⇒ 0 = 5 − 10t
⇒ t = 5/10 s
⇒ t = 0.5 s
Thus, stone will attain a height of 1.25 m and time taken to attain the height is
0.5 s.
Here we have, Acceleration, a = 4 m/s² Initial velocity, u = 0 m/s Time, t = 10 s Distance covered (s) =? We know that, s = ut + 1/2 at² ⇒ s = 0 × 10 + 12 × 4 × (10)² m ⇒ s = 2 × 100 m ⇒ s = 200 m Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration. For more aRead more
Here we have,
Acceleration, a = 4 m/s²
Initial velocity, u = 0 m/s
Time, t = 10 s
Distance covered (s) =?
We know that, s = ut + 1/2 at²
⇒ s = 0 × 10 + 12 × 4 × (10)² m
⇒ s = 2 × 100 m
⇒ s = 200 m
Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration.
Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km /h. What is the average speed for Abdul’s trip?
Let one side distance = x km. Time taken for forward trip at a speed of 20 km/h = Distance / Speed = x/20 h. Time taken in return trip at a speed of 30 km/h = x/30 h. Total time for the whole trip =𝑥/20+𝑥/30=3𝑥+2𝑥/60=5𝑥/60 h. Total distance covered = 2x km. We know, Average speed = Total distance ÷Read more
Let one side distance = x km.
Time taken for forward trip at a speed of 20 km/h = Distance / Speed = x/20 h.
Time taken in return trip at a speed of 30 km/h = x/30 h.
Total time for the whole trip =𝑥/20+𝑥/30=3𝑥+2𝑥/60=5𝑥/60 h.
Total distance covered = 2x km.
We know, Average speed = Total distance ÷ Total time
= 2x ÷ (5x/60) = 24 kmh⁻¹.
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Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
(a) For motion from A to B: Distance covered = 300 m Displacement = 300 m. Time taken = 150 sec. We know that, Average speed = Total distance covered ÷ Total time taken = 300 m ÷ 150 sec = 2 ms⁻¹ Average velocity = Net displacement ÷ time taken = 300 m ÷ 150 sec = 2 ms⁻¹ (b) For motion from A to C:Read more
(a) For motion from A to B:
Distance covered = 300 m
Displacement = 300 m.
Time taken = 150 sec.
We know that, Average speed = Total distance covered ÷ Total time taken
= 300 m ÷ 150 sec = 2 ms⁻¹
Average velocity = Net displacement ÷ time taken
= 300 m ÷ 150 sec = 2 ms⁻¹
(b) For motion from A to C:
Distance covered = 300 + 100 = 400 m.
Displacement = AB – CB = 300 – 100 = 200 m.
Time taken = 2.5 min + 1 min = 3.5 min = 210 sec.
Therefore, Average speed = Total distance covered ÷ Total time taken
= 400 ÷ 210 = 1.90 ms⁻¹.
Average velocity = Net displacement ÷ time taken
= 200 m ÷ 210 sec = 0.952ms⁻¹.
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An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. In 40 sec the athlete complete one round. So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round. ⇒ Distance covered in 140 sec = 2πr × 3.5 = 2 × 22/7 × 100 × 3.5 = 2200 m. At the end of his motion, the athlete will be in the diametricRead more
Time taken = 2 min 20 sec = 140 sec.
Radius, r = 100 m.
In 40 sec the athlete complete one round.
So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.
⇒ Distance covered in 140 sec = 2πr × 3.5 = 2 × 22/7 × 100 × 3.5 = 2200 m.
At the end of his motion, the athlete will be in the diametrically opposite position.
⇒ Displacement = diameter = 200 m.
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A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Here we have, Initial velocity (u) = 5 m/s Final velocity (v) = 0 m/s Acceleration (a) = – 10 m/s² Height, i.e. Distance, s =? Time (t) taken to reach the height =? We know that, v² = u² + 2as ⇒ 0 = (5) ² + 2 × −10 × s ⇒ 0 = 25 − 20s ⇒ s = 25/20 m ⇒ s = 1.25 m Now, we know that, v = u + at ⇒ 0 = 5 +Read more
Here we have,
Initial velocity (u) = 5 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = – 10 m/s²
Height, i.e. Distance, s =?
Time (t) taken to reach the height =?
We know that, v² = u² + 2as
⇒ 0 = (5) ² + 2 × −10 × s
⇒ 0 = 25 − 20s
⇒ s = 25/20 m
⇒ s = 1.25 m
Now, we know that, v = u + at
⇒ 0 = 5 + (–10) × t
⇒ 0 = 5 − 10t
⇒ t = 5/10 s
⇒ t = 0.5 s
Thus, stone will attain a height of 1.25 m and time taken to attain the height is
0.5 s.
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A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Here we have, Acceleration, a = 4 m/s² Initial velocity, u = 0 m/s Time, t = 10 s Distance covered (s) =? We know that, s = ut + 1/2 at² ⇒ s = 0 × 10 + 12 × 4 × (10)² m ⇒ s = 2 × 100 m ⇒ s = 200 m Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration. For more aRead more
Here we have,
Acceleration, a = 4 m/s²
Initial velocity, u = 0 m/s
Time, t = 10 s
Distance covered (s) =?
We know that, s = ut + 1/2 at²
⇒ s = 0 × 10 + 12 × 4 × (10)² m
⇒ s = 2 × 100 m
⇒ s = 200 m
Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-8/