a dragster starting from rest accer;ates at 49m/s how fast is it going when it has raveled 325m?
distance = (1/2) a t^2
that gives you t
then
v = a t
To determine how fast the dragster is going when it has traveled 325m, we can use the following equation of motion:
v^2 = u^2 + 2a s
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s as the dragster starts from rest)
a = acceleration (49 m/s^2)
s = distance traveled (325 m)
Let's substitute the given values into the equation and solve for v:
v^2 = (0 m/s)^2 + 2(49 m/s^2)(325 m)
v^2 = 0 m^2/s^2 + 2(49 m/s^2)(325 m)
v^2 = 0 + 2(49 m/s^2)(325 m)
v^2 = 0 + 2(49 m/s^2)(325 m)
v^2 = 0 + 2(49)(325) m^2/s^2
v^2 = 0 + 31,850 m^2/s^2
v^2 = 31,850 m^2/s^2
Now, take the square root of both sides to find v:
v = sqrt(31,850 m^2/s^2)
v ≈ 178.3 m/s
Therefore, the dragster is going approximately 178.3 m/s when it has traveled 325 m.
To calculate the final speed of the dragster, we can use the kinematic equation:
\(v^2 = u^2 + 2as\),
where:
- \(v\) is the final velocity (speed) of the dragster,
- \(u\) is the initial velocity (which is zero since the dragster starts from rest),
- \(a\) is the acceleration of the dragster, and
- \(s\) is the distance traveled by the dragster.
In this case, the acceleration (\(a\)) of the dragster is given as 49 m/s, and the distance traveled (\(s\)) is 325 m. We need to solve for the final speed (\(v\)).
Rearranging the equation, we have:
\(v^2 = 0^2 + 2 \times 49 \times 325\),
\(v^2 = 0 + 31850\),
\(v^2 = 31850\).
Taking the square root of both sides, we get:
\(v = \sqrt{31850}\).
Calculating \(\sqrt{31850}\), we find that the dragster is going approximately 178.6 m/s (rounded to one decimal place) when it has traveled 325 m.
So, the dragster is going approximately 178.6 m/s when it has traveled 325 m.