A car initially moving at 15 m/s accelerates at 4.5 m/s2 over a distance of 10 m. Find the final velocity?
Let the initial velocity be V1 = 15 m/s and the final velocity be V2 (unknown).
Since F = M a is the accelerating force and the work done is
W = F*X,
K.E. change = (M/2)[V2^2 - V1^2] = M*a*X
M cancels out.
V^2^2 - V1^2 = 2 a X
Solve for V2.
There is another way to derive the same equation without using the mass and force.
Travel time = T = X/(avg velocity)
T = 2X/(V1 +V2)
V2 -V1 = a*T = 2 a X/(V1 + V2)
V2^2 - V1^2 = 2 a X
t=10m/15m.s
=2/3s.
V=u+a¤t
v=(15m/s)+(4.5m/s2)(2/3sec)
v=18m/s
To find the final velocity, you can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance.
The equation is:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Given:
u = 15 m/s (initial velocity)
a = 4.5 m/s^2 (acceleration)
s = 10 m (distance)
Let's substitute the given values into the equation and solve for v:
v^2 = (15 m/s)^2 + 2 * 4.5 m/s^2 * 10 m
v^2 = 225 m^2/s^2 + 90 m^2/s^2
v^2 = 315 m^2/s^2
To find v, we can take the square root of both sides:
v = √(315 m^2/s^2)
v ≈ 17.75 m/s (rounded to two decimal places)
Therefore, the final velocity of the car is approximately 17.75 m/s.