A drag racer starts her car from rest and accelerates at 9.0 m/s2for the entire distance of 60.0 m. What is its speed at the end of the run? Answer in units of m/s.

32.86

s = 1/2 at^2

60 = 4.5t^2
t^2 = 13.333
t = 3.65

v = at = 9*3.65 = 32.86

To find the speed at the end of the run, we can use the kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity (which is 0 m/s since the car starts from rest)
a = acceleration (9.0 m/s^2)
s = distance (60.0 m)

Substituting the given values into the equation:

v^2 = 0^2 + 2(9.0 m/s^2)(60.0 m)
v^2 = 0 + 2(9.0 m/s^2)(60.0 m)
v^2 = 2(9.0 m/s^2)(60.0 m)
v^2 = 1080.0 m^2/s^2

Taking the square root of both sides to find v:

v = āˆš(1080.0 m^2/s^2)
v = 32.91 m/s

Therefore, the speed of the drag racer at the end of the run is approximately 32.91 m/s.

To find the final speed of the drag racer, we can use the equation for uniformly accelerated motion:

v^2 = u^2 + 2as

where
v is the final velocity,
u is the initial velocity (0 m/s, as the car starts from rest),
a is the acceleration (9.0 m/s^2),
and s is the distance traveled (60.0 m).

Since the car starts from rest (u = 0), the equation simplifies to:

v^2 = 2as

Plugging in the given values:

v^2 = 2 * 9.0 m/s^2 * 60.0 m

Now we can calculate the value of v by evaluating the right-hand side of the equation:

v^2 = 1080 m^2/s^2

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

v = āˆš(1080 m^2/s^2)

v ā‰ˆ 32.91 m/s

Therefore, the speed of the drag racer at the end of the run is approximately 32.91 m/s.