Need help with simple acceleration problems.

[ksaechao]

Here's one:

Craig got up to 133.5 km/h getting around a slower vehicle within the length of the passing lane. He then spotted a police cruiser coming around the corner and hit the brakes, slowing to 100 km/h in 0.9 seconds. How much distance did he travel in this time

So I'm pretty much confused with accleration problems and identifying intial velocity and final velocity. I kow that t=.9 secs. So is the initial velocity 133.5 km/h and final velocity is 100 km/h?

Would I first use the delta s=vi(t)+ 1/2(a)t^2 equation to find the displacement of when he slowed down?

 

[mjsd]

Here's one:

Craig got up to 133.5 km/h getting around a slower vehicle within the length of the passing lane. He then spotted a police cruiser coming around the corner and hit the brakes, slowing to 100 km/h in 0.9 seconds. How much distance did he travel in this time

So I'm pretty much confused with accleration problems and identifying intial velocity and final velocity. I kow that t=.9 secs. So is the initial velocity 133.5 km/h and final velocity is 100 km/h?

yes.

Would I first use the delta s=vi(t)+ 1/2(a)t^2 equation to find the displacement of when he slowed down?

to use this you need to assume that acceleration is constant. at this stage you have only 3 known variables: u, v and t. but from these you can work out the average (de)acceleration

 

[m2003]

I would approach this problem by first identifying the given information and what is being asked. In this case, we are given the initial velocity (133.5 km/h), final velocity (100 km/h), and time (0.9 seconds). We are being asked to find the distance traveled during this time.

Next, I would use the formula for acceleration (a) to solve for the acceleration of the vehicle. The formula for acceleration is a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time. Plugging in the given values, we get a = (100 km/h - 133.5 km/h)/0.9 s = -37.5 km/h/s.

Since we now know the acceleration, we can use the formula for displacement (s) to solve for the distance traveled. The formula for displacement is s = vi(t) + 1/2(a)t^2, where vi is the initial velocity, t is the time, and a is the acceleration. Plugging in the given values, we get s = (133.5 km/h)(0.9 s) + 1/2(-37.5 km/h/s)(0.9 s)^2 = 119.25 km.

Therefore, Craig traveled 119.25 km during the 0.9 seconds it took for him to slow down from 133.5 km/h to 100 km/h. It is important to clearly identify and use the correct formulas in solving acceleration problems. I hope this helps with your understanding of acceleration and solving similar problems in the future.