A car starting from rest accelerates at a constant rate of 3.0 m/s2 for 8.0 seconds. How far does the car travel during this time?
Apply newton's 2nd equation of motion which goes asS = u×t +1/2×a×(t^2)Where s is distanceu is initial velocitya is the accelerationt is the time of acceleration.In this problem the car starts from rest so u =0.Putting the other quantities in the equation we seeS = 1/2× 3.0× 8.0 × 8.0S = 96m.Special attention must be given to the units. As all the quantities must be first changed into matching units i.e cgs or s.i. units before applying the equation.
Physics acceleration problem, need help!?
I interpret the problem to mean that a person is holding a dollar on the bottom, lets go, and tries to re-catch it. For the person to catch it, the length of the dollar would have to be at least the distance that it falls. d=1/2*a*t^2, for constant acceleration a and time t. The distance that the dollar would fall in 0.25 sec (assuming no friction) is: d= 1/2*9.8*(.25)^2 = 0.31 m = 31 cm
Negative acceleration dilemma!! Disagreement with my physics teacher?
This is great. thank you all for your time. I do have one thing to add though that is what is making it hard to understand. I realize acceleration is a vector as well as what a vector is, but.... So if my initial Direction is N and as im making my trip i decide to stop and drive in reverse I know my velocity would become negative and it seemed to me that "a" would also. I guess what im saying is that I thought acceleration would be a vector according to the object and not its displacement. In other words lets say a helicopter is flying N and then as he returns home he slows and while he is facing S he accelerates N. According to what i understand now he is accelerating positive now, where he was decelerating with a less negative force before his velocity became 0. I know logic and our methods can not always go hand in hand but this seems like it would be more of a hindrance then help.
Consider a frictionless roller coaster. The acceleration of gravity is 9.8 m/s^2.?
Consider the starting point for the roller coaster to be point A, and the top of the loop is point B. The potential energy at point A is given by: U(A) = mּgּh(A) Solving for h(A) gives h(A) = U(A) / (mּg) So we must find U(A) to determine h(A). Since energy is conserved, we know that throughout the ride, U + K must equal U(A), so we must see if there is a point where we have enough information to find both U and K. We do: point B, the top of the loop. The centripetal acceleration must equal the acceleration of gravity at this point in order for the passengers to feel weightless. So we need to find both U(B) and K(B). Finding U(B) is easy: U(B) = mּgּh(B) h(B) is simply 2r, so substitute: U(B) = 2ּmּgּr K(B) is a bit trickier. There are two accelerations at point B: gravity (simply g) and centripetal (given by a = rּω²). For the passengers to feel weightless, they must be in freefall, so the centripetal acceleration must equal g. a = rּω² = g Solve for angular velocity (ω): g = rּω² ω = √(g/r) Convert the angular velocity to tangential velocity: v = rּω v = rּ√(g/r) Now determine the kinetic energy, K(B): K(B) = ½ּmּv² K(B) = ½ּmּ[rּ√(g/r)]² K(B) = ½ּmּr²ּ(g/r) K(B) = ½ּmּrּg Now we can determine U(A). (Remember, due to conservation of energy, U + K = U(A) at all points.): U(A) = U(B) + K(B) U(A) = 2ּmּgּr + ½ּmּrּg U(A) = 2.5ּmּgּr Substitute this for U(A) in the equation for h(A): h(A) = 2.5ּmּgּr / (mּg) h(A) = 2.5r Stop for a moment and ponder the beauty of physics. Everything drops out except for the radius! That means no matter what the mass is, and no matter what "g" is, h(A) will ALWAYS be 2.5r. This is true whether the roller coaster is built on earth, the moon, or Jupiter, and whether it's loaded with 300-pound linebackers, 110-pound cheerleaders, or 15-pound chihuahuas. Anyway, you can now plug in your known values and calculate the value of h(A): h(A) = 2.5(21 m) h(A) = 52.5 m
Physics question: finding velocity!?
Here is the question: Using a simple pulley/rope system, a crewman on an Arcitc expedition is trying to lower a 5.52 kg crate to the bottom of a steep ravine of a height of 27.0 m. The 55.6 kg crewman is being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 15.0 m from above the ground, the crewman slips and the crate immediately accelerates toward the ground dragging the helpless crewman across the ice and toward the edge of the cliff. If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), AT WHAT SPEED WILL THE CRATE HIT THE GROUND? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff. AT WHAT SPEED WILL THE CREWMAN HIT THE BOTTOM OF THE RAVINE? Thank you!!
A ball is thrown upward with an initial velocity of 30 m/s. what is the maximum height that the ball will reach?
All answers I've read just give you an equation and fill it in. I just don't think that helps anyone understand what is being asked. So here goes my attempt at teaching anyone interested.There actually are several ways to calculate this of which I pick the one that is simplest to me.Assuming there is no (significant) air drag no energy is gained or lost; the kinetic energy (Ek) the ball has leaving your hand is converted in potential gravitational energy (Ep) on the way up, and back to kinetic energy on the way down.Their size must therefore be equal:Ek = EpAlso we know that,Ek = 0.5*m*v^2. and, Ep = m*g*hwith m the mass of the ball, v its velocity leaving your hand, g the gravitational acceleration and h the maximum height the ball reaches.Rearanging a bit gives you:v^2 = 2*g*h or,v^2/2g =hCurious about a different way?Take,St = S0 + V0 *t + 0.5*a*t^2which gives the position of an object at time t (St) given the position at time 0 (S0) the initial velocity (V0) and the acceleration (a) which is constant.For a we fill in -g (because down is negative) we take S0 = 0 and we know V0.To find t we need to solve it from,Vt = V0 + a*t.Since the velocity at the summit of the balls flight is zero we get:V0 = g*t. or, t = V0/g(Notice how I cleverly got rid of a minus?)Fill in that value into the equation for St and you're done.But you can also plug the formula for t into the equation:St = V0*V0/g - 0.5*g*(V0/g)^2= (V0^2)/g - 0.5*g*(V0^2)/g^2= (V0^2)/g - 0.5*(V0^2)/g= 0.5*(V0^2)/g = (V0^2)/2gwhich is the result we had before when you consider that v there is V0 here.Isn't it great how everything lines up?I hope this helped you understand rather than get the answer right.Yes… There's a difference…
Physics question, please help.....?
A) One way to solve this is to make use of the following equation: (Vf)^2 = (Vi)^2 + 2ad (Equation 1) Where Vf = the final velocity; Vi = the initial velocity a = acceleration and d = distance. Note, this equation can be derived from the relationships Vf = Vi +at ; where t = time (Equation 2) and d = Vi(t) + 1/2at^2 (Equation 3) Squaring Equation 2 = Vf^2 = Vi^2 +2Vi(at) +(at)^2 or Vf^2 = Vi^2 + 2a[Vi(t) + 1/2at^2] Substituting in Equation 3 gives us Vf^2 = Vi^2 +2ad (Equation 1) 100kilometers per hour = 100,000m/h (100,000m/h)/ 3600s/h = 27.78 m/s = required take off speed Use Equation 1 to solve for Vf when the runway is 150m long (Vf)^2 = (0)^2 +2(2m/s^2)(150m) = 600m^2/s^2 Vf = 24.5m/s This is not fast enough to take off. B) Again use Equation 1 with Vf = 27.78m/s , and solve for d (27.78m/s)^2 = (0)^2 + 2(2m/s^2)d d = (771m^2/s^2) / (4m/s^2) d = 192.9mThus the runway needs to be 192.9m long.