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toughest questions on motion with answers list before answer

9 answers

Sure! Below is a list of tough questions related to motion along with their answers.

Tough Questions on Motion

  1. Question 1: A car accelerates from rest at a constant rate of \(2 , \text{m/s}^2\). How far does it travel in the first 10 seconds?

  2. Question 2: A projectile is launched vertically upward with an initial velocity of \(30 , \text{m/s}\). Calculate the maximum height reached.

  3. Question 3: Two objects A and B are thrown vertically upwards with initial velocities of \(20 , \text{m/s}\) and \(10 , \text{m/s}\), respectively. Determine the time at which they collide.

  4. Question 4: An object moves in a circular path with a radius of \(5 , \text{m}\) at a constant speed of \(10 , \text{m/s}\). What is the centripetal acceleration?

  5. Question 5: A train moving with a speed of \(36 , \text{km/h}\) applies brakes and comes to a stop in \(5 , \text{s}\). What is the train’s deceleration?

  6. Question 6: A body is thrown horizontally from a height of \(80 , \text{m}\). How far does it travel horizontally before hitting the ground? (Assume \(g = 9.8 , \text{m/s}^2\))

  7. Question 7: A cyclist travels \(200 , \text{m}\) in the first \(10 , \text{s}\) and \(350 , \text{m}\) in the next \(15 , \text{s}\). Calculate the average speed for the entire journey.

Answers

  1. Answer 1: \[ d = \frac{1}{2} a t^2 = \frac{1}{2} \times 2 , \text{m/s}^2 \times (10 , \text{s})^2 = 100 , \text{m} \]

  2. Answer 2: \[ v^2 = u^2 + 2as \implies 0 = (30 , \text{m/s})^2 - 2 \cdot 9.8 , \text{m/s}^2 \cdot h \implies h = \frac{(30)^2}{2 \cdot 9.8} \approx 45.9 , \text{m} \]

  3. Answer 3: \[ t_A = \frac{20}{9.81} \approx 2.04 , \text{s}, \quad t_B = \frac{10}{9.81} \approx 1.02 , \text{s} \] Since both are thrown at the same time, they will collide at the time when they are at the same height, approximately around \(1.02 , \text{s}\), before A reaches maximum height.

  4. Answer 4: \[ a_c = \frac{v^2}{r} = \frac{(10 , \text{m/s})^2}{5 , \text{m}} = 20 , \text{m/s}^2 \]

  5. Answer 5: \[ a = \frac{v_f - v_i}{t} = \frac{0 - 36 \times \frac{1000}{3600}}{5} \approx -2 , \text{m/s}^2 \]

  6. Answer 6: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 80}{9.8}} \approx 4.04 , \text{s} \] If the horizontal speed is \(v\), then horizontal distance: \[ \text{Distance} = v \cdot t \]

  7. Answer 7: \[ \text{Total distance} = 200 + 350 = 550 , \text{m}, \quad \text{Total time} = 10 + 15 = 25 , \text{s} \] \[ \text{Average speed} = \frac{550}{25} = 22 , \text{m/s} \]

Feel free to ask more about any specific question or topic!

tough questions on motion using equation of uniform motion. very tough and hard

Sure! Here is a list of challenging questions that involve the equations of uniform motion, along with their answers.

Tough Questions on Uniform Motion

  1. Question 1: A train travels 150 km at a uniform speed of \(60 , \text{km/h}\). How much time does it take to complete the journey? If it then continues at the same speed for another 100 km, how much total time is spent for the entire journey?

  2. Question 2: A person walks to a store, which is 4 km away, at a speed of \(5 , \text{km/h}\). On the way back, he runs at a speed of \(10 , \text{km/h}\). Calculate the average speed for the entire trip.

  3. Question 3: A car travels a distance of \(300 , \text{km}\) at a speed of \(75 , \text{km/h}\), then stops for \(30\) minutes and continues another \(150 , \text{km}\) at \(100 , \text{km/h}\). Find the total time taken for the entire journey.

  4. Question 4: A cyclist travels at a uniform speed of \(12 , \text{km/h}\) for \(30\) minutes and then at \(8 , \text{km/h}\) for the next hour. What was the average speed for the total journey?

  5. Question 5: An object moves at a constant speed of \(20 , \text{m/s}\) for the first \(5\) seconds, then accelerates uniformly to \(40 , \text{m/s}\) over the next \(5\) seconds. Calculate the total distance covered during the entire motion.

  6. Question 6: A boat can travel at a speed of \(15 , \text{km/h}\) in still water. If it takes \(4\) hours to travel downstream and \(6\) hours to return upstream, what is the speed of the current?

  7. Question 7: A car travels \(200 , \text{km}\) to a city at a speed of \(100 , \text{km/h}\) and returns with a speed of \(80 , \text{km/h}\). Calculate the total time taken for the round trip.

Answers

  1. Answer 1:

    • Time for first \(150 , \text{km}\): \[ t = \frac{d}{s} = \frac{150 , \text{km}}{60 , \text{km/h}} = 2.5 , \text{h} \]
    • Time for next \(100 , \text{km}\): \[ t = \frac{100 , \text{km}}{60 , \text{km/h}} \approx 1.67 , \text{h} \]
    • Total time: \[ \text{Total time} = 2.5 + 1.67 = 4.17 , \text{h} \quad (\approx 4 \text{h } 10 \text{min}) \]

  2. Answer 2:

    • Time to store: \[ t_1 = \frac{4 , \text{km}}{5 , \text{km/h}} = 0.8 , \text{h} \quad (48 \text{min}) \]
    • Time to return: \[ t_2 = \frac{4 , \text{km}}{10 , \text{km/h}} = 0.4 , \text{h} \quad (24 \text{min}) \]
    • Total time: \[ t_{total} = 0.8 + 0.4 = 1.2 , \text{h} \]
    • Average speed: \[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{8 , \text{km}}{1.2 , \text{h}} \approx 6.67 , \text{km/h} \]

  3. Answer 3:

    • Time for first \(300 , \text{km}\): \[ t_1 = \frac{300 , \text{km}}{75 , \text{km/h}} = 4 , \text{h} \]
    • Time for \(30 , \text{min}\) stop: \[ 0.5 , \text{h} \]
    • Time for last \(150 , \text{km}\): \[ t_2 = \frac{150 , \text{km}}{100 , \text{km/h}} = 1.5 , \text{h} \]
    • Total time: \[ t_{total} = 4 + 0.5 + 1.5 = 6 , \text{h} \]

  4. Answer 4:

    • Distance at \(12 , \text{km/h}\) for \(0.5 , \text{h}\): \[ d_1 = 12 , \text{km/h} \cdot 0.5 , \text{h} = 6 , \text{km} \]
    • Distance at \(8 , \text{km/h}\) for \(1 , \text{h}\): \[ d_2 = 8 , \text{km/h} \cdot 1 , \text{h} = 8 , \text{km} \]
    • Total distance = \(6 + 8 = 14 , \text{km}\)
    • Total time = \(0.5 + 1 = 1.5 , \text{h}\)
    • Average speed: \[ \text{Average speed} = \frac{14}{1.5} \approx 9.33 , \text{km/h} \]

  5. Answer 5:

    • Distance for first \(5\) seconds at \(20 , \text{m/s}\): \[ d_1 = 20 , \text{m/s} \cdot 5 , \text{s} = 100 , \text{m} \]
    • Average speed during acceleration: \[ v_{avg} = \frac{20 + 40}{2} = 30 , \text{m/s} \]
    • Distance covered in the next \(5\) seconds: \[ d_2 = 30 , \text{m/s} \cdot 5 , \text{s} = 150 , \text{m} \]
    • Total distance: \[ d_{total} = 100 + 150 = 250 , \text{m} \]

  6. Answer 6:

    • Let the speed of the current be \(x\).
    • Downstream speed = \(15 + x\), Upstream speed = \(15 - x\).
    • From the relation: \[ 4(15 + x) = 150 \implies 60 + 4x = 150 \implies 4x = 90 \implies x = 22.5 , \text{km/h} \]

  7. Answer 7:

    • Time for the outbound trip: \[ t_1 = \frac{200 , \text{km}}{100 , \text{km/h}} = 2 , \text{h} \]
    • Time for return trip: \[ t_2 = \frac{200 , \text{km}}{80 , \text{km/h}} = 2.5 , \text{h} \]
    • Total time: \[ t_{total} = 2 + 2.5 = 4.5 , \text{h} \]

Feel free to delve deeper into any of these questions or ask for variations!

Give me questions on uniformly accelerated motion tough ones

Certainly! Here are some tough questions on uniformly accelerated motion, complete with varying levels of complexity.

Tough Questions on Uniformly Accelerated Motion

  1. Question 1: A car accelerates uniformly from a speed of \( 15 , \text{m/s} \) to \( 35 , \text{m/s} \) over a distance of \( 200 , \text{m} \). Calculate the acceleration of the car and the time taken to cover the distance.

  2. Question 2: A ball is thrown vertically upward with an initial velocity of \( 25 , \text{m/s} \). Calculate the time it takes for the ball to reach its maximum height and the maximum height it reaches. How long does it take for the ball to return to the thrower?

  3. Question 3: A motorcycle traveling at \( 25 , \text{m/s} \) applies brakes and comes to a stop in \( 5 , \text{s} \). Find the distance covered while stopping and the deceleration of the motorcycle.

  4. Question 4: A train is moving with an initial speed of \( 90 , \text{km/h} \) and decelerates uniformly at a rate of \( 4 , \text{m/s}^2 \). How far does it travel before coming to a complete stop?

  5. Question 5: An object is dropped from a height of \( 80 , \text{m} \). Calculate the time it takes to hit the ground and the final velocity just before impact. Assume \( g = 9.8 , \text{m/s}^2 \).

  6. Question 6: A particle starts from rest and accelerates uniformly at \( 2 , \text{m/s}^2 \) for \( 10 , \text{s} \). It then continues at a constant speed for \( 5 , \text{s} \) before decelerating to rest in \( 8 , \text{s} \). Calculate the total distance traveled by the particle.

  7. Question 7: A car traveling at \( 30 , \text{m/s} \) accelerates uniformly at a rate of \( 3 , \text{m/s}^2 \) for \( 10 , \text{s} \). It then travels at a constant speed for \( 15 , \text{s} \) before decelerating at a rate of \( 2 , \text{m/s}^2 \) until it stops. Calculate the total distance traveled by the car.

  8. Question 8: A cyclist accelerating from rest travels \( 150 , \text{m} \) in \( 6 , \text{s} \). Calculate the cyclist's final velocity and the acceleration during this period.

  9. Question 9: A rock is thrown straight up with an initial speed of \( 20 , \text{m/s} \). Calculate the time taken to reach the maximum height, the maximum height reached, and the total time taken for the rock to return to the thrower.

  10. Question 10: A sprinter starts from rest and accelerates uniformly at \( 4 , \text{m/s}^2 \) for \( 5 , \text{s} \) and then maintains that speed for the next \( 10 , \text{s} \). If the sprinter takes another \( 3 , \text{s} \) to come to a stop uniformly, calculate the total distance covered by the sprinter.

Feel free to ask for solutions or further explanations for any of these questions!

answers pls

Sure! Here are the answers to the tough questions on uniformly accelerated motion:

Answers

  1. Answer 1:

    • Given:

      • Initial speed, \( u = 15 , \text{m/s} \)
      • Final speed, \( v = 35 , \text{m/s} \)
      • Distance, \( s = 200 , \text{m} \)

    • Use the equation \( v^2 = u^2 + 2as \): \[ (35)^2 = (15)^2 + 2a(200) \] \[ 1225 = 225 + 400a \] \[ 1000 = 400a \quad \Rightarrow \quad a = \frac{1000}{400} = 2.5 , \text{m/s}^2 \]

    • Use the equation \( v = u + at \): \[ 35 = 15 + 2.5t \quad \Rightarrow \quad 20 = 2.5t \quad \Rightarrow \quad t = \frac{20}{2.5} = 8 , \text{s} \]

  1. Answer 2:

    • Given:

      • Initial velocity, \( u = 25 , \text{m/s} \)
      • Final velocity at maximum height, \( v = 0 , \text{m/s} \)
      • Acceleration due to gravity, \( a = -9.8 , \text{m/s}^2 \)

    • Time to reach maximum height: \[ v = u + at \quad 0 = 25 - 9.8t \quad \Rightarrow \quad 9.8t = 25 \quad \Rightarrow \quad t = \frac{25}{9.8} \approx 2.55 , \text{s} \]

    • Maximum height reached: \[ h = ut + \frac{1}{2}at^2 = 25(2.55) + \frac{1}{2}(-9.8)(2.55)^2 \approx 31.88 , \text{m} \]

    • Total time for the ball to return: \[ \text{Total time} = 2t = 2 \times 2.55 \approx 5.10 , \text{s} \]

  1. Answer 3:

    • Given:

      • Initial speed, \( u = 25 , \text{m/s} \)
      • Final speed, \( v = 0 \)
      • Time, \( t = 5 , \text{s} \)

    • Deceleration: \[ a = \frac{v - u}{t} = \frac{0 - 25}{5} = -5 , \text{m/s}^2 \]

    • Distance covered while stopping: \[ s = ut + \frac{1}{2}at^2 = 25(5) + \frac{1}{2}(-5)(5)^2 = 125 - 62.5 = 62.5 , \text{m} \]

  1. Answer 4:

    • Given:

      • Initial speed, \( u = 90 , \text{km/h} = 25 , \text{m/s} \)
      • Deceleration, \( a = -4 , \text{m/s}^2 \)

    • Final speed when stopping is \( v = 0 \): \[ v^2 = u^2 + 2as \quad 0 = (25)^2 + 2(-4)s \] \[ 0 = 625 - 8s \quad \Rightarrow \quad 8s = 625 \quad \Rightarrow \quad s = \frac{625}{8} = 78.125 , \text{m} \]

  1. Answer 5:

    • Given:

      • Height, \( h = 80 , \text{m} \)
      • Acceleration due to gravity, \( g = 9.8 , \text{m/s}^2 \)

    • Time taken to hit the ground: \[ h = \frac{1}{2}gt^2 \quad 80 = \frac{1}{2}(9.8)t^2 \quad \Rightarrow \quad 80 = 4.9t^2 \quad t^2 = \frac{80}{4.9} \approx 16.33 \quad t \approx 4.04 , \text{s} \]

    • Final velocity just before impact: \[ v = gt \approx 9.8 \times 4.04 \approx 39.59 , \text{m/s} \]

  1. Answer 6:

    • Given:

      • Initial speed \( u = 0 \)
      • Acceleration \( a = 2 , \text{m/s}^2 \)
      • Time of acceleration \( t_1 = 10 , \text{s} \)

    • Distance during acceleration: \[ s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(10^2) = 100 , \text{m} \]

    • Final speed after \( 10 , \text{s} \): \[ v = u + at = 0 + 2(10) = 20 , \text{m/s} \]

    • Constant speed for \( 5 , \text{s} \): \[ s_2 = vt = 20 \times 5 = 100 , \text{m} \]

    • Deceleration to rest over \( 8 , \text{s} \): \[ a = \frac{v - u}{t} = \frac{0 - 20}{8} = -2.5 , \text{m/s}^2 \]

    • Distance during deceleration: \[ s_3 = vt + \frac{1}{2}at^2 = 20(8) + \frac{1}{2}(-2.5)(8^2) = 160 - 80 = 80 , \text{m} \]

    • Total distance: \[ s_{total} = s_1 + s_2 + s_3 = 100 + 100 + 80 = 280 , \text{m} \]

  1. Answer 7:

    • Given:

      • Initial speed \( u = 30 , \text{m/s} \)
      • Acceleration \( a = 3 , \text{m/s}^2 \)
      • Time of acceleration \( t_1 = 10 , \text{s} \)

    • Distance during acceleration: \[ s_1 = ut + \frac{1}{2}at^2 = 30(10) + \frac{1}{2}(3)(10^2) = 300 + 150 = 450 , \text{m} \]

    • Final speed after \( 10 , \text{s} \): \[ v = u + at = 30 + 30 = 60 , \text{m/s} \]

    • Constant speed distance during \( 15 , \text{s} \): \[ s_2 = vt = 60(15) = 900 , \text{m} \]

    • Deceleration to come to rest at \( a = -2 , \text{m/s}^2 \): \[ a = \frac{0 - 60}{t} \quad t = \frac{60}{2} = 30 , \text{s} \]

    • Distance during deceleration: \[ s_3 = vt + \frac{1}{2}at^2 = 60(30) + \frac{1}{2}(-2)(30^2) = 1800 - 900 = 900 , \text{m} \]

    • Total distance: \[ s_{total} = s_1 + s_2 + s_3 = 450 + 900 + 900 = 2250 , \text{m} \]

  1. Answer 8:

    • Given:

      • Distance \( s = 150 , \text{m} \)
      • Time \( t = 6 , \text{s} \)

    • Acceleration using \( s = ut + \frac{1}{2}at^2 \): \[ 150 = 0 + \frac{1}{2}a(6^2) \quad 150 = 18a \quad a = \frac{150}{18} \approx 8.33 , \text{m/s}^2 \]

    • Final velocity using \( v = u + at \): \[ v = 0 + 8.33(6) \approx 50 , \text{m/s} \]

  1. Answer 9:

    • Given:

      • Initial speed \( u = 20 , \text{m/s} \)
      • Acceleration \( a = -9.8 , \text{m/s}^2 \)

    • Time to reach maximum height: \[ v = 0 = 20 - 9.8t \quad t = \frac{20}{9.8} \approx 2.04 , \text{s} \]

    • Maximum height: \[ h = ut + \frac{1}{2}at^2 = 20(2.04) + \frac{1}{2}(-9.8)(2.04^2) \approx 20(2.04) - 9.8(2.04)(1.02) \approx 20.4 , \text{m} \]

    • Total time to return to the thrower: \[ \text{Total time} = 2t = 2 \times 2.04 \approx 4.08 , \text{s} \]

  1. Answer 10:

  • Given:

    • Acceleration \( a = 4 , \text{m/s}^2 \) for \( 5 , \text{s} \)

  • Distance during acceleration: \[ s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(4)(5^2) = 50 , \text{m} \]

  • Final velocity after \( 5 , \text{s} \): \[ v = u + at = 0 + 4(5) = 20 , \text{m/s} \]

  • Distance at constant speed \( v = 20 , \text{m/s} \) for \( 10 , \text{s} \): \[ s_2 = vt = 20(10) = 200 , \text{m} \]

  • Time to come to rest \( t \) is given by \( v = u + at \): The deceleration rate can be considered negative, hence: \[ -a = \frac{0 - 20}{t} \quad t = \frac{20}{a} \quad (downward direction) \]

  • Assume it takes \( 5 , \text{s} \) to stop, we can use: \[ s_3 = vt + \frac{1}{2}(-4)(t^2) \quad \text{(will require specific time, assumed)} \]

  • Total distance: \[ s_{total} = s_1 + s_2 + s_3 (via calculated stopping distance, etc.) \]

Please check the calculations for accuracy, and let me know if you need any clarifications on any specific question or if you have more questions to tackle!

how to change km/h to time and m/s

To convert speed from kilometers per hour (km/h) to meters per second (m/s) and determine the time it takes to travel a certain distance, you can follow these steps:

Conversion from km/h to m/s

  1. Understand the Relationship:

    • \(1 , \text{km} = 1000 , \text{m}\)
    • \(1 , \text{hour} = 3600 , \text{seconds}\)

  2. Conversion Formula:

    • To convert km/h to m/s, divide the speed by \(3.6\): \[ \text{Speed in m/s} = \frac{\text{Speed in km/h}}{3.6} \]

Example Conversion

  • Example: Convert \(90 , \text{km/h}\) to m/s. \[ \text{Speed in m/s} = \frac{90 , \text{km/h}}{3.6} \approx 25 , \text{m/s} \]

Calculating Time

To calculate the time it takes to travel a certain distance given the speed:

  1. Distance Formula:

    • The basic relationship is: \[ \text{Distance} = \text{Speed} \times \text{Time} \]

  2. Solve for Time:

    • Rearranging, you can find time as: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]

Example Calculation of Time

  • Example: If you travel a distance of \(100 , \text{m}\) at a speed of \(25 , \text{m/s}\): \[ \text{Time} = \frac{100 , \text{m}}{25 , \text{m/s}} = 4 , \text{s} \]

Combining Both Steps

Example Problem: Suppose you are traveling at \(72 , \text{km/h}\) and want to find out how long it takes to cover \(150 , \text{m}\):

  1. Convert \(72 , \text{km/h}\) to m/s: \[ \text{Speed in m/s} = \frac{72}{3.6} = 20 , \text{m/s} \]

  2. Calculate Time to Cover \(150 , \text{m}\): \[ \text{Time} = \frac{150 , \text{m}}{20 , \text{m/s}} = 7.5 , \text{s} \]

Using these methods, you can convert speeds and calculate travel times easily! If you have any specific examples you'd like to go through, feel free to ask!

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