A football player can accelerate at 2 m/s^2. if he starts from rest. how long will it take him to travel 16 m?
It is a simple problem on the uniformly accelerated rectilinear motion. The hourly equation is: s=1/2 a t^2+vot+s0 vo is the initial speed that is zero because the football player starts from rest s0 is the initial position of football player that you can take equal to zero 16=0.5 2 t^2 t=4 sec
A ball starts moving with an acceleration of 4m/s2. What is the speed after 4 seconds?
Assuming constant acceleration,Velocity will vary linearly , by definition of acceleration;a = change in velocity / timeOr we can say ; change in velocity = acceleration× timeFinal velocity - initial velocity = 4×4=16 m/s( Since ball start with an acceleration of 4m/s2 , so initial velocity is assumed zero )Final velocity= 16 m/sSpeed after 4 second will be 16m/s
A car accelerates from rest with 2m/s2 on a straight line and then comes to a rest. The total distance traveled by the car is 100 m in 20 sec. What is the maximum velocity attained by the car?
V (1) [initial velocity] = zeroLet V (2) represent the max velocity, which is what we need to determine.V (f) [Final velocity] = zeroLet d (1) represent the distance covered till V (2) and t (1) the time taken.Let d (2) represent the retardation distance and t (2) the retardation time. Then;d (2) = 100 - d (1) [EQ 1] and t (2) = 20 - t (1) [EQ 2]We have five variables, and we need five equations and when we solve the five equations simultaneously, we obtain a value for V (2). We set up our five equations using kinematics.For the first leg of the trip: d (1) = 1/2 (a) [t (1)]^2 → d (1) = t (1)^2 [EQ 3]V (2) = a t (1) + 0 = 2 t (1) [EQ 4]For the last leg of the trip, the distance is the product of the average velocity and time: d (2) = [0 + V (2)] [t (2)] / 2 → d (2) = t (2) * [V (2) / 2] {EQ 5}Solving:Substitute [3] in [1] → d (2) = 100 - t^2 [EQ 6]Substitute [6] in [5] → 100 - t (1)^2 = [V (2) * t (2)] / 2 [EQ 7]Substitute [4] in [2] → t(2) = 20 - [V (2) /2)] [EQ 8]Substitute [4] & [8] in [7] → 100 - [V (2) / 2]^2 = [V (2) / 2] [20 - {V (2)/2}]Remove parentheses and solve: 100 - V^2 / 4 = 10 V (2) - V^2 / 410 V (2) = 100V (2) = 10 m/s QEDCHECK: at 2 m/s/s, it takes the car 5 seconds to reach 10 m/s. The car covers 25 meters during that time. The second half of the journey has a distance of 75 meters and using average velocity we obtain time to be 15 seconds. the sum of 5 and 15 = 20 seconds. CHECK.
Help w/ PHYSICS QUESTION! Two soccer players start from rest, 51 m apart...?
(df- di) (change in distance) = .5at^2 + (t)vo (initial velocity) a1 = .50 vo= o d= .5(.5)t^2 = .25t^2 a2 = .38 vo=0 df= .5(-.38)t^2 + 51 (This is di) = .19t^2+51 one acceleration must be negative compared to the other. set them equal to each other. .25t^2 = -.19t^2 + 51 .44t^2 = 51 t= 10.77 seconds
A quarterback throws a football toward a receiver with an initial speed of 25 m/s at an angle of 30° above the?
y= displacement of y= 0 m (because it ends on the same level of the y axis at which it started) v0= initial velocity= 25 m/s angle= 30 degrees v0y= initial velocity in terms of y= v0sin(angle)= 25sin30 = 12.5 m/s v0x= initial velocity in terms of x= v0cos(angle)= 25cos30= 21.65 m/s a= acceleration= -9.8 m/s (due to gravity) t= time= ? First step is to solve for time. We will use this formula: y=v0y*t + (1/2)*a*t^2 so, subbing in the information we know for the given variables: 0= 12.5t + (1/2)*(-9.8)*t^2 Divide both sides by t 0= 12.5 + (1/2)(-9.8)t -12.5= (1/2)(-9.8)t -12.5= -4.9t t=2.55 seconds Step two is to find how far the displacement is in the x direction given t=2.55 seconds so v0x is the initial velocity of the ball in the x direction, so...: vox*t= x 21.65(2.55)= x x= 55.2 m Now we know that the ball is going to travel 55.2 m before it is at the same level that it was kicked at. Our receiver is already 15 m away from the quarter back so 55.2- 15 = 40.2 m is how far he will need to run in order to catch the ball before it hits the ground, in 2.55 seconds. So to find how fast he will need to run, x, we will use the formula x= distance/time so x= 40.2/2.55 x= 15.76 m/s
Physics help-displacement/vectors, acceleration, velocity...?
1. Both vectors have a y component of 100 km, therefore if you add them together the total y component is 100 km + 100 km = 200 km. We know that the angle to the x-axis is 60° so we can form the equation: y = r.sin(60) where y = the total y component = 200 km and r = the total magnitude 200 = r.sin(60) 200 = (r.√3)/2 r = 400/√3 = (400.√3)/3 = 231 km (3sf) 2. His total distance is 4 km + 8 km = 12 km, his total time = 0.4 hr + 0.8 hr = 1.2 hr 12 km / 1.2 hr = 10 km/hr 3. Use the equation of motion: a = Δv/t where: a = average acceleration Δv = change in velocity = 20 m/s - 0 = 20 ms t = time = 8 s a = 20/8 = 2.5 m/s² 4. Use the equation of motion s = u.t + ½.a.t² where: s = displacement u = initial velocity = 0 a = acceleration = 5 m/s² t = time = 4 s s = 0×4 + ½×5×4² = 40 m but he started 10 m from the goal so he ends up 10 m + 40 m = 50 m away
A ball is dropped from a height of 45m. What will be the time to reach the ground?
The distance is 45m. (s)Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)We have to find the time (t)By the 3rd equation of motion,s = ut + 1/2 at^2 (here u is the initial velocity of the ball)Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.(substituting the values in the formula)45 = 1/2 x 9.8 x t^245 = 9.8/2 x t^245 = 4.9 x t^2 (9.8/2 = 4.9)(divide 4.9 on both sides)45/4.9 = 4.9 x t^2 / 4.99.1836 = t^23.030 = tSo it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).
A football was kicked at 22.5 m/s at an angle of 35 degrees above the horizontal. What was the ball's hangtime?
22.5m/s at 35° means the vertical ball speed (ascending) is sin(35°) x 22.5m/s = 12.9m/s.The gravity force is now accelerating the ball downward at 9.81m/s². We can deduce the time the ball will stop going up and start going down by doing 12.9/9.81 = 1.315 s.Now we know it takes about 1.315 second for the ball to stop going up. As the acceleration rate is constant, the ball will take the same time to reach the ground.So the total flight time of the ball will be about 2 x 1.315 = 2.63 seconds.Of course this result do not include height difference between the starting point and the landing point, neither the energy loss due to friction with air. To take friction into account, we will need the mass and the diameter of the ball.
A ball is thrown vertically upward with a speed of 20 m/s. When will it reach the maximum height? What is the maximum height reached?
These questions can be answered by making use of Newton's equations of motion. There are 3 equations of motions.[math]v = u + at[/math][math]s = ut + \frac{1}{2}at^2[/math][math]v^2 = u^2 + 2as[/math]Where,v = final velocityu = initial velocitya = accelerationt = times = distanceIn your question, the initial velocity is given as [math]20 m/s[/math], i.e., [math]u = 20 m/s[/math], the final velocity that the ball can achieve at the maximum height is [math]0 m/s[/math], hence, [math]v = 0 m/s[/math]. Since the only first that cause the acceleration is gravity, a is taken as g where g is acceleration due to gravity, and had a value of [math]9.81 m/s^2[/math]. But for simplicity, we can take the value of a to be [math]10 m/s^2[/math], so [math]a = 10 m/s^2[/math]. Now, we need to find, what's s and t.Note: Since the ball is thrown upwards, which is against the force of gravity (gravity always acts downwards), we need take the value of a (in this case, g) as [math]-10 m/s^2[/math]. Using the first equation,[math]v = u + at[/math][math]0 = 20 - 10t[/math][math]10t = 20[/math][math]t = 2[/math]Using the third equation,[math]v^2 = u^2 + 2as[/math][math]0^2 = 20^2 + 2 × (-10) × s[/math][math]20s = 400[/math][math]s = 20[/math]Hence, the ball will travel for 2 seconds and complete a distance of 20 metres upwards.
A ball is thrown vertically upwards. What is the final velocity and acceleration at the maximum height?
This is physics at its most “common sense” form!You just need to think about you throwing a ball in the air. When you throw a ball, it’s only logical that at some point it stops and then comes back down to you. So, the velocity at the maximum height(the point where it turns around) is zero!Now, the acceleration is a bit more tricky but quite easy too! Which is the force that tries to bring the ball back to you*? Well, it’s the force that tries to keep you on the ground; it’s dear old gravity! But, does it change depending on where the ball is located? No. And we know that the gravitational acceleration is approximately 9.8 m/s^2 and, as I said, it’s constant. So, at maximum height, and at any height, the acceleration of the ball is equal to the gravitational acceleration!I honestly think that you should have thought about this much harder before you posted it as a question in Quora; this is the way to build intuition. You first start from simple, intuitive things and build onward from them to things that are a bit unintuitive at first. Cheers!*In my answer, I am not considering any force from the air around the ball, of course.