Help with physics?!?

Question

28. Calculate the acceleration of a car that can go from rest to 100 km/hr in 10 seconds.

>>You have to converst the seconds to hours, right? I got its acceleration was 30000 something.... and that's totally impractical. What do I do to convert them properly?

#36. Which has more acceleration--a car increasing its speed from 50 km/h to 60 km/h or a bicycle that goes from zero to 10 km/h in the same time? Defend your answer.

>>Wouldn't the acceleration be the same?

#41. A ball is thrown straight up. What will be the instantaneous velocity at the top of its path? What will its acceleration be at the top? Why are your answers different?

>>The instantaneous velocity is zero, right? And wouldn't the acceleration be zero, as well?

#48. a) Find the speed required to throw a ball straight up and have it return 6 seconds later. Neglect air resistance.

>>I got to v=V-instantaneous - 10 m/s^2(6s). I have two variables though--what would I put in for v so I could solve for V-instantaneous?

#48. b) How high does the ball go?

>> I could solve this once I had the speed, right? If so, then I can do it on my own. If not... well, I'd rather like some help.

#50. If a salmon swims straight upward in the water fast enough to break through the surface at a speed of 5 m/s, how can it jump above the water?

>>I feel like its really straightforward and I should know how to do it, but I can't. Do I need more information?

If you can show me how to do these without actually giving me the answer, I'd prefer that.I'd rather understand and do it on my own than just get the answer. However, if you can't... well, I'll let you decide if you want to give me the answer or not.

Thanks so much!

Answer 1

for 28, you know v(0) = 0m/s and v(10s) = (100km/hr x hr/3600s x 1000m/km) <-- that's your conversion. From there plug it into the equation v(t) = a(t) + v(0)t, when t=10 (and at t=10, v(t) = what you sdolved for above in m/s)

36. good answer, but back up the reason by using the slope: the change in the velocity is the same

41. Acceleration stays constant: as the ball slows down in the positive direction, the change in the slope of the velocity is negative. However, instantaneous velocity is the slope of the tangent of the position graph, so yes, it would be 0

48. a) unless you know V(0) (or v-not), then you can't solve for it except through an equation with variables

b) use the position equation

50. it's velocity is positive in the y-direction, and gravity accelerates it in the downward direction, but it takes time for gravity to cause it's velocity to reach 0.

Answer 2

lol sorry im in ap math i dont get that see if an carcalater will work

Answer 3

28.  100 km/hr * (1000m/1km) * (1 hr/60 min)* (1 min/60sec) = 100/3.6 m/s (divide by time to get accel)

Looks like the accel should be in the neighborhood of 3 m/s/s

36. Yes, same change in speed over change in time

41.  Yes v=0, No a≠0, a=-g (in free flight the accel is the acceleration due to gravity, that's why it comes back down.  If it had accel =0 and v=0, then the velocity would stay 0 and there would be no motion and it would just hang there up in the air).

48. hint: 3 seconds up, and 3 seconds down.  Or you could fill in the blanks for this equation  Ã¢ÂˆÂ†y=vo t - 0.5gt^2, where t=6 sec, g=9.8 m/s/s and ∆y=0, solve for vo.

48 again.  You could use v^2=vo^2-2g∆y, where v=velocity at top of flight which you already said you knew. vo from the first part.

#50 not sure what exactly you are looking for, it starts out moving vertically with an initial speed of 5 m/s, that carries it up beyond the surface of the water.

Good luck!!! Hope this helps.