Practice 2D: | 1 | 2
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**Velocity and displacement with uniform acceleration -
by Kevin Bailey & Chicken Breast, 2002**

**1. A car with an initial speed of 23.7 km/h accelerates
at a uniform rate of 0.92 m/s ^{2} for 3.6 s. Find the final speed
and the displacement of the car during this time.**

Here's what you know, 23.7 km/h = 6.58 m/s. Use the formula v

_{f}= v_{i}+ a(Dt) for the final speed and plug in: 6.58 m/s = the initial speed, 0.92 m/s^{2}= the acceleration, and 3.6 s = the time. v_{f}= (6.58 m/s) + (0.92 m/s^{2})(3.6 s) =9.89 m/s or 36 km/h... fast car.Now, let's find the displacement! (Hey you might as well get excited) Use the formula Dx = v

_{i}(Dt) +^{1}/_{2}a(Dt)^{2}and simply plug in the variables. That gives you Dx = (6.58 m/s)(3.6 s) +^{1}/_{2}(0.92 m/s^{2})(3.6 s)^{2}=29.6 m or 0.030 km

**2. An automobile with an initial speed of 4.30 m/s
accelerates uniformly at the rate of 3.0 m/s ^{2}. Find the final
speed and the displacement after 5.0 s.**

** **Ok, so
first to solve for the displacement. You need to know this formula. Dx
= v_{i}(Dt) + ^{1}/_{2}a(Dt)^{2},
where:

Dx = ???

v_{i} = 4.30 m/s

Dt = 5.0 s

a = 3.0 m/s^{2}

So, Dx
= (4.30 m/s)(5 s) + (^{1}/_{2})(3.0 m/s^{2})(5.0 s)^{2}
= **59 m**

**
**Got it? Good. Alrighty then, to solve for the speed this is all
you need: v_{f} = v_{i} + a(Dt),
v_{f} = ?, v_{i} = 4.30 m/s, a = 3.0 m/s^{2}, Dt
= 5.0 s

So that's v_{f} = (4.30
m/s) + (3.0 m/s^{2})(5.0 s) = **19.3 m/s**

**3. A car starts from rest and travels for 5.0 s with a
uniform acceleration of -1.5 m/s ^{2}. What is the final velocity
of the car? How far does the car travel in this time interval?**

Well, well, velocity and displacement again...but there's a catch. This time the car is starting from rest. We will use the same equations as we did in the last two, except this time substitute in 0 for the initial velocity.

This leaves us with the speed being v

_{f}= v_{i}+ a(Dt) or v_{f}= (0 m/s) + (-1.5 m/s^{2})(5.0 s) =-7.5 m/sI wonder why the driver is backing up so quickly?So then, we are back to the displacement. Again use Dx = v

_{i}(Dt) +^{1}/_{2}a(Dt)^{2}and just substitute in the variables. Dx = (0)(5.0 s) +^{1}/_{2}(-1.5 m/s^{2})(5.0 s)^{2}=-18.75 m or -19 m(sig figs).

**4. A driver of a car traveling at 15.0 m/s applies the
brakes, causing a uniform acceleration of -2.0 m/s ^{2}. How long
does it take the car to accelerate to a final speed of 10.0 m/s? How far
has the car moved during the braking period?**

Aw geez, the last one. Why did the fun have to end so soon? But hey we're in luck, because this time we get to do some more algebra!

Ok, given initial and final velocities and the acceleration we must find the time. In order to do this simply use the same formula for the final speed and put your math skills to work. So v

_{f}= v_{i}+ a(Dt) and 10 m/s = (15 m/s) + (-2.0 m/s^{2})(Dt). Then we are left with -5 m/s = (-2.0 m/s^{2})(Dt) and Dt = (-5 m/s) / (-2.0 m/s^{2}) =2.5 sNow then, let's see how far this particular automobile traveled. Again use Dx = v

_{i}(Dt) +^{1}/_{2}a(Dt)^{2}and just substitute in the variables. Dx = (15 m/s)(2.5 s) +^{1}/_{2}(-2.0 m/s^{2})(2.5 s)^{2}=31.25 m or 31 m

TA DAAA!!