Help on this mechanics question please [suvat equations]?
A particle is moving with constant acceleration in a straight line. It passes through three points, A, B and C with speeds 20ms^-1, 30ms^-1 and 45ms^-1 respectively. The time taken to move from A to B is t1 seconds and the time taken to move from B to C is t2 seconds. a) Show that t1/t2 = 2/3 Given also that the total time taken for the particle to move from A to C is 50s, b) find the distance between A + B. Thanksssssssssssssssss + please explain how to do it !
How do you approach and answer these type of mechanic questions?
In the next two seconds, it'll travel 14m further. Let me tell you why... You know that, s = u.t + (1/2).a.(t^2) where s is the displacement of the moving object; u is the velocity of the object when it started moving; t is time, and a is acceleration. According to the question, the object travels 10m in the first two seconds and 12m in further 2 seconds. So you can conclude that the object travels 22m in the first 4 seconds. Using the given data we can form two equations with two variables, i.e, u (primary velocity) and a (acceleration). The equations are as follow... 10 = 2.u + (1/2).a.(2^2) ---------(i) 22 = 4.u + (1/2).a.(4^2) ---------(ii) Solving the above equations you will know both the primary velocity as well as the acceleration of the object. a = 0.5 m/s^2; u = 4.5 m/s. Hence you can easily find how far it'll travel in the next 2 seconds. i.e, in 6 seconds. In 6 seconds it'll travel 36m. So, it'll travel (36 - 22)m further. i.e, 14m further.
Projectile question ? helpp please?
The gravitational constant on earth is 9.81m/s. That's how fast they accelerate. Now I'm not completely sure if this'll work but it's worth a shot. SUVAT is the most helpful tool in Mechanics S = Distance (Measured in meters) (40) U = Starting Velocity (Measured in m/s) (0) V = Ending Velocity (Measured in m/s) (Unknown) A = Acceleration (Measured in m/s2 (That's meters per second squared) (9.81) T = Time (Measured in s) (Unknown) As long as you have at least 3 of the 5 values you can work out the rest with these equations: v = u + at s = ut + (1/2)(a)(t^2) s = vt - (1/2)(a)(t^2) v^2 = u^2 + 2as s = ((u + v)/2)t Right we first want to know V, We have to choose an equation that DOESN'T have T in it. V squared = U squared + 2 x A x S There we can do that one. Fill in the gaps: 0x0+2x9.81x40=784.8 The square root of 784.8 = 28.014 So V = 28.014 Now we find T. V=U+AT V-U=AT U=0 so V=AT 28.014=9.81xT V/A=T 28.014/9.81=T 28.014/9.81=2.856 So T = 2.856 Noticed anything yet? If it takes 2.856 seconds for the anvil to hit the bottom of the cliff then you times 2.856 by 10. So thats 28.56. Which is the answer. That's why Roadrunner has to be 28.56 meters away before Wile.E.Coyote drops the anvil, because it takes 2.856 seconds to reach the bottom! Although the answer's rounded off. So the answer is 28.6 m. Like you said. :D You're welcome. (Oh and don't forget, all objects fall and accelerate at the same speed regardless of mass or weight, Don't forget that, this is why we don't need the weight of the anvil to work this out.) I don't think the answer has taken into account that Roadrunner is 1 meter tall. You would do this with 39 meters otherwise but I did it with 40 as an example and it's easier to work with. Also I knew it would give the right answer. These details are normally overlooked. Even by the question asker himself most of the time.
An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 135km/h .?
Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.20m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)? I have worked this problem numerous times and every time I get it wrong. Could someone clearly explain how to work this problem? Thank you in advance.
The work done in accelerating an object along a frictionless horizontal surface is equal to the?
Because it changes from potential energy to kinetic energy and the change requires force to happen. The object has to go from stationary to moving, so the work equals the energy exerted to cause the movement/change.
How do I solve the given laws of motion problem for the JEE Advanced?
TO solve all these types of questions there is one simple method which is applicable on all questions of this type1.Find the tension in the strings connected to the blocks.2.Multiply the total tension at each block to their velocity.3.Take a sign convention for ex-Upward motion=positiveDownward motion=negative4.Sum all the constraints related to each block and equate them to zero5.You will easily find a relation between the speed of blocks.i would like to tell that i also used to get in trouble due to these type of problems because the technique of differentiation given in most books is very complex and takes a lot of time .But after knowing this trick i can solve many questions just by seeing.
In free-fall acceleration, how do I figure when "g" is positive or negative?
g is a constant defined as the magnitude of free-fall acceleration (acceleration of an object under the influence of gravity alone). As a magnitude, g is always positive. However, I suspect what you are asking is how to determine whether the direction of the acceleration vector in free-fall is positive or negative.As mentioned in other answers, free-fall acceleration is a vector quantity that always points toward the center of the earth. Any object moving away from the earth under the influence of gravity alone will decrease in velocity until it reaches max height where velocity becomes zero, at which point it will move back toward the earth, increasing in velocity until it reaches earth's surface. The difference between any two successive velocity vectors in such a situation will always point toward earth. Therefore, the acceleration vector due to gravity (difference in velocity over time) will always point toward earth as well.When performing calculations involving g, however, an important distinction must be made between acceleration due to gravity (a vector) and the constant g (the magnitude of the acceleration vector). Whether or not your acceleration vector is positive or negative depends on your choice of coordinate system. If you choose up to be positive (as is commonly done for ballistics or free-fall problems), then the direction of the vector is going to be negative. However, regardless of your choice of origin or coordinate system, the value of g will always be positive.