Тест. Равноускоренно прямолинейное движение. Ускорение (практика)

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Список вопросов теста

Вопрос 1

Автомобиль начинает движение из состояния покоя. Через 50 с он достигает скорости 72 км/ч. С каким ускорением происходит движение?

Варианты ответов

• 0,4 м/с²

• 72 м/с²

• 0,8 м/с²

• -0,4 м/с²

90000 Uniformly Accelerated Motion Calculator 90001 90002 Calculator Use 90003 90004 The Uniformly Accelerated Motion calculator uses the equations of motion to solve motion calculations involving constant acceleration in one dimension, a straight line. It can solve for the initial velocity u, final velocity v, displacement s, acceleration a, and time t. 90005 90004 Choose a calculation to find the variables that are unknown and enter the variables that are given in your problem.This calculator will calculate the unknown values ​​and provide the derived equations that were used to find the solution. Solution equations are derived from the uniformly accelerated motion equations below. 90005 90004 Note that when solving for multiple variables there is usually more than one way to solve for your unknowns. You can derive more than one set of equations to solve your problem in different ways. 90005 90010 Uniformly Accelerated Motion Equations 90003 90004 \ (s = \ dfrac {1} {2} (v + u) t \ tag {1} \) 90005 90004 \ (v = u + at \ tag {2} \) 90005 90004 \ (v ^ 2 = u ^ 2 + 2as \ tag {3} \) 90005 90004 \ (s = ut + \ dfrac {1} {2} at ^ 2 \ tag {4} \) 90005 90004 Where: 90005 90022 90023 u = initial velocity 90024 90023 v = final velocity 90024 90023 a = acceleration 90024 90023 s = displacement 90024 90023 t = time 90024 90033 90004 Use standard gravity, a = 9.80665 m / s 90035 2 90036, for equations involving the Earth’s gravitational force as the acceleration rate of an object. 90005 90004 Equations 1 through 4 are the key equations used to solve for variables in this calculator however you will sometimes see a different number of Uniformly Accelerated Motion Equations depending on the resource. You will find that equation 1 comes from substituting equation 1b into equation 1a below. 90005 90004 \ (s = \ overline {v} t \ tag {1a} \) 90005 90004 \ (\ overline {v} = \ dfrac {1} {2} (v + u) \ tag {1b} \) 90005 .90000 Uniformly Accelerated Motion — IB Physics Stuff 90001 90002 Definition of IB symbols: 90003 90002 90005 u 90006 initial velocity 90007 90005 v 90006 final velocity 90007 90005 a 90006 acceleration (constant) 90007 90005 t 90006 time taken 90007 90005 s 90006 distance traveled 90003 90020 90005 2.1.7 Derive the equations for uniformly Accelerated Motion 90006 90023 90002 Acceleration is defined as the rate of change of velocity with respect to time, in a given direction.The SI units of acceleration are ms 90025 -2 90026. This would mean that if an object has an acceleration of 1 ms 90025 -2 90026 it will increase its velocity (in a given direction) 1 ms 90025 -1 90026 every second that it accelerates. If we write the definition for acceleration in mathematical terms: 90003 (1) 90002 \ begin {align} a = {v-u \ over t} \ end {align} 90003 90002 Then solve of v: 90003 (2) 90002 \ begin {equation} v = u + at \ end {equation} 90003 90002 This gives us the final velocity of an object in terms of the acceleration, initial velocity and the time the object was accelerating.In graphical form the equation is a straight line, with u as the vertical intercept, a as the slope and t as the independent variable. 90003 90002 We can derive yet another equation if we look at the graph of a velocity vs. time for an object with a non-zero initial velocity. 90003 90002 We know that the area under the curve is equal to the distance travele 90003 (3) 90002 \ begin {equation} Area = Red + Yellow \ end {equation} 90003 (4) 90002 \ begin {align} s = ut + {(v-u) \ over 2} t \ end {align} 90003 (5) 90002 \ begin {align} s = ut + {1 \ over 2} vt — 1 {\ over 2} ut \ end {align} 90003 (6) 90002 \ begin {align} s = {u + v \ over 2} t \ end {align} 90003 90002 In we now combine the two equations that we have derived, we can create another equation: 90003 (7) 90002 \ begin {align} s = {u + u + at \ over 2} t \ end {align} 90003 (8) 90002 \ begin {align} s = ut + {1 \ over 2} at ^ 2 \ end {align} 90003 90002 Now if we go back to the definition of acceleration (1) and multiply by t and divide by a: 90003 (9) 90002 \ begin {align} t = {v-u \ over a} \ end {align} 90003 90002 If we substitute this expression for time into the 3rd equation and solve for v: 90003 (10) 90002 \ begin {align} s = u \ left [{v-u \ over a} \ right] + {1 \ over 2} a \ left [{v-u \ over a} \ right] ^ 2 \ end {align} 90003 (11) 90002 \ begin {align} s = {uv -u ^ 2 \ over a} + {v ^ 2 + u ^ 2 -2uv \ over 2a} \ end {align} 90003 (12) 90002 \ begin {equation} 2as = 2uv -2u ^ 2 + v ^ 2 + u ^ 2 -2uv \ end {equation} 90003 (13) 90002 \ begin {equation} 2as = v ^ 2 -u ^ 2 \ end {equation} 90003 (14) 90002 \ begin {equation} v ^ 2 = u ^ 2 + 2as \ end {equation} 90003 90002 So what do we have? We have four equations that describe uniformly accelerated motion: 90003 90002 Equation (2) provides a way of calculating the final velocity in terms of the initial velocity, acceleration and the time the object was accelerated.90003 90002 Equation (6) provides a way of calculating the distance (displacement) of an object in terms of the initial velocity, final velocity and time the object was in motion. 90003 90002 Equation (8) gives us the distance traveled without having to know the final velocity of the object. In exchange for knowing the final velocity we must know the acceleration of the object. 90003 90002 Equation (14) relates the initial velocity, final velocity and acceleration of the object without time! Sometimes this is very useful.90003 90002 It is very important to note that 90005 these equations ONLY apply if the object in question is experiencing uniform acceleration 90006, meaning that the acceleration is constant or can be approximated as constant. 90003 90020 90005 2.1.8 Describe the vertical motion of an object in a uniform gravitational field 90006 90023 90002 A uniform gravitational field simply means that the force of gravity does not change or does not change significantly. If you stay near the surface of the Earth (with in a kilometer or two) the gravitational field can be assumed to be constant (at least for the purposes of IB physics).90003 90002 If the gravitational field is constant then an object in that field will experience a constant force and thus a constant acceleration. This means that the equations derived above are a valid description of the motion of an object in a uniform gravitational field (assuming no other forces). 90003 90002 If an object is held stationary in a uniform gravitational field it will fall. It will do so with uniform acceleration. Near the surface of the earth the acceleration is approximately 9.8 ms 90025 -2 90026. This means that every second that the object falls its velocity will increase by 9.8 ms 90025 -1 90026. So after one second the object has a velocity of 9.8 ms 90025 -1 90026, after 5 seconds it will have a velocity of 47.5 ms 90025 -1 90026, etc. Since the objects velocity is increasing every second this naturally means that the distance it is covering each second also increases … 90003 90002 What happens if an object is thrown up? The acceleration is still downward. If an object is thrown up with an initial velocity of 30 ms 90025 -1 90026, after one second it will only be going 20 ms 90025 -1 90026 up, after 2 seconds it will only be going 10 ms 90025 -1 90026, after 3 seconds the object will have zero velocity! Even if the objects velocity is zero the acceleration is not zero.90003 90002 Vertical motion in a gravitational field will be discussed at more length when we get to projectile motion. 90003 90020 90005 2.1.9 Describe the effects of air resistance of a falling object 90006 90023 90002 As an object falls it experiences a drag or frictional force due to air, we call this force air resistance. This drag force is always in the opposite direction of the motion. When the object is moving slowly the force of drag is proportional to the speed of the object. As the object’s speed increase the drag is proportional to the square of the speed, which means that the drag force becomes large very quickly.At some point the drag force will increase to the size of the force of gravity on the object. When the two forces are equal their will be no net force on the object and it will no longer accelerate, i.e. it will now travel at a constant velocity. This final maximum velocity is called the terminal velocity. The drag and thus the terminal velocity is determined by the shape of the object, its mass and the cross section of the object. 90003 90002 People have survived falling out of airplanes and hot air balloons, this does not happen often but it does happen once in a great while.They many fall for a couple of minutes before hitting the ground, where they actually bounce. It happens that a falling human body will reach its terminal velocity (about 200-300 km / hr) fairly quickly, so while they may fall for a long period of time they hit the ground at a relatively low speed. Low compared to the speed they would hit with if the air did not reduce their acceleration. 90003 90124 90125 Want to add to or make a comment on these notes? Do it below. 90126 .90000 Equations of Constant Acceleration Motion 90001 90002 The 90003 uniformly accelerated rectilinear motion (u.a.r.m.) 90004, also known as 90003 constant acceleration motion 90004, is a rectilinear motion that has a constant acceleration, which is different from zero. In this section we are going to study: 90007 90008 Concept of constant acceleration motion 90009 90002 The 90003 constant acceleration motion 90004 is quite common in your daily life. An object that is allowed to fall and that does not find any obstacle in its way (free fall), or a skier that descends an incline, just before arriving to the jump area, are good examples of this.The constant acceleration motion or 90003 uniformly accelerated rectilinear motion (u.a.r.m) 90004 has the following 90003 properties 90004: 90007 90002 A body moves with 90003 constant acceleration motion 90004 or 90003 uniformly accelerated rectilinear motion (u.a.r.m) 90004 when its 90023 90003 trajectory is a straight line 90004 90026 and its 90023 90003 acceleration is constant and different from 0 90004 90026. This implies that 90023 the velocity increases or decreases its magnitude uniformly 90026.90007 90034 90035 Uniformly Accelerated Rectilinear Motion 90007 90002 In our example the car describes a u.a.r.m since it moves in a straight line with a constant acceleration equivalent to 2 m / s 90038 2 90039. [Notice that, in each second, the velocity and the distance traveled by the body increase based on the value of the acceleration in the previous second.] 90007 90002 Notice that although colloquially, we make distinction between an accelerating and a braking body, from the point of view of physics, both are 90003 uniformly accelerated rectilinear motions 90004.The difference is that while one has positive acceleration, the other one has negative acceleration. 90007 90008 Constant acceleration motion equations 90009 90002 The 90003 equations of the constant acceleration motion 90004 or 90003 uniformly accelerated rectilinear motion (u.a.r.m.) 90004 are: 90007 90002 Where: 90007 90055 90056 90023 x 90026, 90023 x 90060 0 90061 90026: Position of the body at a given time (90023 x 90026) and at the initial time (90023 x 90060 0 90061 90026).Its unit in the International System (SI) is the meter (m) 90069 90056 90023 v 90026, 90023 v 90060 0 90061 90026: 90003 90023 Velocity 90026 90004 of the body at a given time (90023 v 90026) and at the initial time (90023 v 90060 0 90061 90026). Its unit in the International System is meter per second (m / s) 90069 90056 90023 a 90026: 90023 90003 Acceleration 90004 90026 of the body. Remains constant with a value different from 0. Its unit in the International System is meter per second (m / s 90038 2 90039) 90069 90056 90023 t 90026: 90003 90023 Time 90026 90004 being studied.Its unit in the International System is the second (90023 s 90026) 90069 90108 90002 Although the former are the main equations of the u.a.r.m. and the only ones necessary to solve the exercises, it is sometimes useful know the following expression: 90007 90002 v2 = v02 + 2 · a · Δx 90007 90002 The above formula allows you to relate velocity and distance traveled if the acceleration is known, and can be deduced from the previous ones, as you can see below. 90007 90002 v = v0 + a · tx = x0 + v0 · t + 12 · a · t2⇒t = v-v0aΔx = v0 · t + 12 · a · t2⇒Δx = v0v-v0a + 12 · a · v-v0a2; 90007 90002 2 · a · Δx = v2-v02 90007 90119 Deduction of the constant acceleration motion equations 90120 90002 To deduce the 90003 equations of constant acceleration motion 90004 or 90003 uniformly accelerated rectilinear movement (u.a.r.m) 90004, it must be taken into consideration that: 90007 90055 90056 The normal or centripetal acceleration value is zero: an = 0 90069 90056 The average acceleration, instant acceleration and tangential acceleration have the same value: a = aa = at = cst 90069 90108 90002 With these restrictions, we get: 90007 90002 aa = aaa = ΔvΔt = v-v0t-t0 = ⏟t0 = 0x-x0t → v-v0 = a⋅t → v = v0 + a⋅t 90007.90000 Uniformly Accelerated Motion MCQs — Quiz Questions and Answers 90001 90002 Uniformly accelerated motion MCQs, uniformly accelerated motion quiz answers pdf to study online physics degree course. Learn applied physics: motion and force Multiple Choice Questions and Answers (MCQs), «uniformly accelerated motion» quiz questions and answers for online bachelor degree programs. Learn displacement in physics, velocity formula, momentum in physics, physics equations test prep for ACT test. 90003 90002 «At surface of the Earth where air friction is negligible, objects fall with the» Multiple Choice Questions (MCQ) on uniformly accelerated motion with choices 90005 different acceleration, same velocity, same acceleration, and same speed 90006 for online bachelor degree programs .Practice merit scholarships assessment test, online learning uniformly accelerated motion quiz questions for competitive exams in physics majors for SAT test prep classes. Uniformly Accelerated Motion Video 90003 90008 MCQs on Uniformly Accelerated Motion PDF Book Download 90009 90002 MCQ: At surface of the Earth where air friction is negligible, objects fall with the 90003 90012 90013 different acceleration 90014 90013 same velocity 90014 90013 same acceleration 90014 90013 same speed 90014 90021 90002 MCQ: From the equations of distance, correct one is 90003 90012 90013 V 90026 f 90027 = V 90026 i 90027 + 2as 90014 90013 V 90026 f 90027 ² = V 90026 i 90027 ² + as 90014 90013 V 90026 f 90027 ² = V 90026 i 90027 ² + 2a 90014 90013 V 90026 f 90027 ² = V 90026 i 90027 ² + 2as 90014 90021 90002 MCQ: Correct equation of distance is 90003 90012 90013 V 90026 i 90027 = V 90026 f 90027 + at 90014 90013 V 90026 f 90027 = V 90026 i 90027 + at 90014 90013 V 90026 f 90027 = V 90026 i 90027 + t 90014 90013 V 90026 f 90027 = V 90026 i 90027 + a 90014 90021 90002 MCQ: Velocity of the vertically thrown ball with time will be 90003 90012 90013 upwards positive 90014 90013 downwards negative 90014 90013 upwards negative 90014 90013 downwards positive 90014 90021 .