Motion Problems?

Question

d = d0 + v0t + ½ a t2
v = v0t + a t
v2 = v02 + 2ad

A car accelerates from 5 m/sec to 10 m/sec with an acceleration of 1 m/sec/sec.
a. How far did the car travel during this acceleration?

I don't need the answer to this problem, I would really just like how to get this problem going. The equations on top are the ones I need to use and I have no idea where to start. Any help is great.

Answer 1

Simplest method: note that the car is accelerating for 5 seconds. Therefore, find the distance it travels starting from zero and 5m/s in 5 seconds. That is, d=5t+t²/2 (d_0=0, v_0=5, a=1). This gives you d=25+12.5=37.5 meters

Answer 2

The first equation has everything you need. d = the total distance which is what you are looking for. d0 = the initial starting point. In this case you don't care because the problem only asks for the distance during the acceleration period, so d0 can = 0. V0 is the velocity at time t0 and V0 is given as 5m/sec. Acceleration= "a" is given as 1 m/sec/sec. t is easy to calculate in your head as 5 sec. So now just plug and chug.
d = 0 + 5*5 + 1/2*1*5^2 = 37.5 m

It was easy to figure t=5sec in this problem, but if the numbers had been harder, then you could calculate it from the second equation but I think it should read v = v0 + at. It means the eventual velocity = the initial velocity + a*t. So 10m/sec= 5m/sec + 1m/sec/sec times t. So t = 5m/sec.

Answer 3

Safest way is to use the equations given. Let's see, the information you KNOW (ie. the givens), are your initial velocity (v0=5), final velocity (v=10), and your acceleration (a=1). Given that, the only equation you can work with (to find distance) is the last one: velocity squared equals the initial velocity squared plus double the product of the acceleration and distance travelled.

Thus, d = (v^2 - v0^2)/2a = (100 - 25)/2 = 75/2 = 37.5 meters.

Also, the second equation should be v = v0 + at, not v = v0t + at. Acceleration is the derivative (rate of change) of Velocity, so the derivative of v should be a. So if y = mx + b, m= a (a is the slope of the velocity). v = at + v0 is equivolent to y = mx + b (your initial velocity is the y intercept of your graph, since at zero you start with y equalling the initial velocity).

Answer 4

Don't know all the fancy equations for this, but . . .

5 to 10 m/sec at 1m/sec/sec takes 5 sec.

second 1 = 5 m
second 2 = 6 m
second 3 = 7 m
second 4 = 8 m
second 5 = 9 m

5+6+7+8+9=35 m.

Answer 5

acceleration is the rate of cange od speed therefore
a = du /dt ....we will need the time taken for accelerating form 5m.s to 10 m/se =>

dt= (t) = 10-5 / 1 = 5 sec.

now you treat the problem as a normal problem of body that accelerated with an known a for a known t with initial velocity (not from rest)


so s = Uinital* t + 1/2 a t^2 = 5 * 5 + 1/2 * 1 * 5^2 = 37.5m

Answer 6

Your first eq is OK, but the other 2 are garbage. The second one should read v = v0 + at

From this, put in your numbers: 10 = 5 + 1*t ---> t = 5 sec

Then from eq 1), d = 0 + 5*5 + (1/2)*1*5^2 = 25 + 12.5 = 37.5m

Answer 7

the question is asking for distance so definately use the third equation. initial and final v's are given as well as acceleration, so just solve for distance.

http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1c.html trying checking that out...i actually read that when i was taking physics. also physicsforums.com is a great help!

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