1)Top-fuel drag racers are able to uniformly accelerate at 12.3 m/s2 from rest to 98 m/s before crossing the finish line. How much time elapses during the run?

2)A racecar accelerates from rest at +7.2 m/s2 for 4.1 seconds. How fast will it be going at the end of that time?

3)A bullet accelerates at 6.8 X 10^4 m/s2 from rest as it travels the 0.80 m of a rifle barrel. What is the velocity of the bullet as it leaves the barrel?

v = 0 + a*t

98 = 12.3 * t
solve for t

v = 0 + a*t again

x = X0 + Vo t +.5 a t^2
.8 = 0 + 0 + (.5) (6.8*10^4 )t^2
t^2 = 2.35 10^-5
t = .00485 seconds in barrel
then use
v = o + a t again

A bullet accelerates at 6.8*10^4 m/s^2 from rest as it travels the 0.80 m of the rifle barrel. How long was then bullet in the barrel? What velocity does the bullet have as it leaves the barrel?

1) To find the time elapsed during the run, we can use the equation:

v = u + at

Where:
v = final velocity (98 m/s)
u = initial velocity (0 m/s)
a = acceleration (12.3 m/s^2)
t = time

Rearranging the equation to solve for time:

t = (v - u) / a

Substituting the given values:

t = (98 m/s - 0 m/s) / 12.3 m/s^2
t = 98 m/s / 12.3 m/s^2
t ≈ 7.97 seconds

Therefore, approximately 7.97 seconds elapse during the run.

2) To find the final speed of the racecar, we can use the equation:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity (0 m/s)
a = acceleration (+7.2 m/s^2)
t = time (4.1 seconds)

Substituting the given values:

v = 0 m/s + 7.2 m/s^2 * 4.1 s
v = 0 m/s + 29.52 m/s
v ≈ 29.52 m/s

Therefore, the racecar will be going approximately 29.52 m/s at the end of 4.1 seconds.

3) To find the velocity of the bullet as it leaves the barrel, we can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (unknown)
u = initial velocity (0 m/s)
a = acceleration (6.8 x 10^4 m/s^2)
s = distance traveled (0.80 m)

Substituting the given values:

v^2 = 0 m/s + 2 * (6.8 x 10^4 m/s^2) * 0.80 m
v^2 = 0 m^2/s^2 + 13.6 x 10^4 m^2/s^2
v^2 = 13.6 x 10^4 m^2/s^2

To find the velocity, we take the square root of both sides:

v = √(13.6 x 10^4 m^2/s^2)
v ≈ 369.08 m/s

Therefore, the velocity of the bullet as it leaves the barrel is approximately 369.08 m/s.

To solve these problems, we will use the equations of motion. The key equation we need is:

v = u + at

where:
v = final velocity
u = initial velocity (usually 0 in these cases since the objects start from rest)
a = acceleration
t = time

1) To find the time it takes for the top-fuel drag racer to reach a velocity of 98 m/s, we can rearrange the equation:

v = u + at
98 = 0 + (12.3)t

Now, solve for t:

t = 98 / 12.3
t ≈ 7.98 seconds

Therefore, the time it takes for the racer to reach a velocity of 98 m/s is approximately 7.98 seconds.

2) To find the final velocity of the racecar after accelerating at 7.2 m/s^2 for 4.1 seconds, we can again use the equation:

v = u + at
v = 0 + (7.2)(4.1)

Now, calculate the final velocity:

v = 7.2 * 4.1
v ≈ 29.52 m/s

Therefore, the racecar will be going at approximately 29.52 m/s at the end of 4.1 seconds.

3) To calculate the velocity of the bullet as it leaves the rifle barrel, we can use the same equation:

v = u + at

The bullet starts from rest (u = 0) and the acceleration is given as 6.8 x 10^4 m/s^2. The distance traveled is 0.80 m. We do not know the time, so let's rearrange the equation:

v = u + at
v = 0 + (6.8 x 10^4) t

Now, we need to find the time. To do this, we will use another equation of motion:

s = ut + (1/2)at^2

We want to find the time when the bullet travels a distance of 0.80 m:

0.80 = 0 + (1/2)(6.8 x 10^4)t^2

Simplify the equation:

0.80 = 3.4 x 10^4t^2

Now solve for t:

t^2 = 0.80 / (3.4 x 10^4)
t ≈ √(0.80 / 3.4 x 10^4)

Once you calculate t, substitute it back into the first equation:

v = (6.8 x 10^4) t

Calculate the velocity:

v ≈ (6.8 x 10^4) * (t value)

This will give you the velocity of the bullet as it leaves the barrel.

Keep in mind that when solving problems like these, it is important to pay attention to units and use appropriate unit conversions if necessary.