Calculating Motion Equations for a Sprinter and Jet Aircraft | Homework Help

[Sins]

Motion Equation Help, Please :)

Hi, could anyone help me out with the following? My teacher gave us homework without explaining it :'(. I've tried to work out the answers, but I don't know if they're right.

So, could someone work them out for me, please?
This question is about a sprinter running a 100m race.
(a) A sprinter accelerates uniformly to a maximum speed over the first 30m of a race in a time of 5.0s. Calculate his acceleration.
(I've tried this one, but my answer isn't a very logical one.) I got 6 m.s-2 for this one. Don't know if it's right though.

(b) Calculate his maximum velocity.
(I can't do this one, at all. I don't even know how to calculate velocity. Could someone please explain how to do this? Thanks, could you also give me the equation to doing this one?)

(c) He then travels at uniform velocity a further 70m to the finish line. Calculate the time taken for the 70m and hence his total time for the 100m race.

(2)A jet aircraft has an acceleration of 6m.s-2. If initially the aircraft is traveling at 80 m.s-1, how long will it take to accelerate to 330 m.s-1 (the speed of sound)?

Thanks for the help :D

 

[HallsofIvy]

What, your teacher expects you to think for yourself? Why how awful of him/her!

(a) A sprinter accelerates uniformly to a maximum speed over the first 30m of a race in a time of 5.0s. Calculate his acceleration.

You probably have an equation in your book that says if something moves with constant velocity a from an initial velocity 0, then the distance he goes in t seconds is d(t)= (1/2)at². You are told that when t= 5s, d= 30m. Put those into the equation and solve for a.

(a) A sprinter accelerates uniformly to a maximum speed over the first 30m of a race in a time of 5.0s. Calculate his acceleration.

acceleration is defined as "change in speed divided by change in time" or
a= v/t. That is the same as v= at. You just found a in the first part of the problem and you know that time is t= 5s. (In this problem, since he starts with speed 0, "change in speed" is the same as "maximum speed.)

(c) He then travels at uniform velocity a further 70m to the finish line. Calculate the time taken for the 70m and hence his total time for the 100m race.

You just found his speed after the acceleration in part (b). That the "uniform velocity" mentioned. You should know that "velocity is distance divided by time" or v= d/t. You know v and d= 70. Put those into the equation and solve for t. Don't forget to add the original 5 seconds accelerating time to get the total time for the 100 m.

(2)A jet aircraft has an acceleration of 6m.s-2. If initially the aircraft is traveling at 80 m.s-1, how long will it take to accelerate to 330 m.s-1 (the speed of sound)?

Again, acceleration equals change in speed divided by time. If the jet was originally traveling at 80 m/s and goes up to 330 m/s how much has its speed changed? You are told that the acceleration is 6 m/s². Put those into the equation and solve for t.

 

[quicknote]

Hi there! I'm happy to help with your homework on calculating motion equations for a sprinter and jet aircraft. Let's start with the sprinter question:

a) To calculate acceleration, we use the equation a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. In this case, we know that the sprinter's initial velocity (vi) is 0 m/s, because they start from a standstill. We also know that the final velocity (vf) is the maximum speed they reach, and the time (t) is 5.0 seconds. So, plugging these values into the equation, we get: a = (vf - 0)/5.0 = vf/5.0. Since we don't know the final velocity yet, we can't solve for acceleration. However, we do know that the sprinter reaches a maximum speed over the first 30m, so we can use the equation vf^2 = vi^2 + 2ad (where d is distance) to solve for vf. Plugging in the values we know, we get: vf^2 = (0)^2 + 2(6)(30) = 360. This means that the final velocity is the square root of 360, which is approximately 18.97 m/s. Now, we can plug this value into our original equation to solve for acceleration: a = (18.97)/5.0 = 3.79 m/s^2. So, the sprinter's acceleration is 3.79 m/s^2.

b) To calculate maximum velocity, we can use the same equation as before: vf^2 = vi^2 + 2ad. We know that the sprinter's initial velocity (vi) is 0 m/s, and we just solved for acceleration (a = 3.79 m/s^2). We also know that the distance (d) is 30m. So, plugging in these values, we get: vf^2 = (0)^2 + 2(3.79)(30) = 227.4. This means that the maximum velocity is the square root of 227.4, which is approximately 15.08 m/s. So, the sprinter's maximum velocity is 15.08 m/s.

c) Since the sprinter travels at uniform