A truck covers 40.0 m in 7.00 s while smoothly slowing down to a final velocity of 2.45 m/s.
Find the truck's original speed.
Find its acceleration.
nope
This has already been answered. See your previous post.
To find the truck's original speed, we can use the equation of motion:
v = u + at
where:
v = final velocity = 2.45 m/s
u = initial velocity (original speed) - what we need to find
a = acceleration - what we will find later
t = time = 7.00 s
We also know that the distance covered is 40.0 m.
We can use the first equation of motion to find the acceleration:
s = ut + (1/2)at^2
where:
s = distance = 40.0 m
u = initial velocity = what we need to find
a = acceleration - what we will find
t = time = 7.00 s
Now let's solve these equations step by step:
1. From the first equation of motion:
v = u + at
2.45 = u + a(7.00)
2. Rearrange the equation to solve for u:
u = 2.45 - 7.00a ----(Equation 1)
3. Plug this value of u into the second equation of motion:
s = ut + (1/2)at^2
40.0 = (2.45 - 7.00a)(7.00) + (1/2)a(7.00)^2
4. Simplify and solve for a:
40.0 = 17.15 - 49.00a + 1/2 * 49.00a
40.0 = 17.15 - 24.50a
5. Rearrange the equation to solve for a:
24.50a = 17.15 - 40.0
24.50a = -22.85
a = -22.85 / 24.50
a ≈ -0.93 m/s^2
6. Now, substitute the value of a into Equation 1 to find u:
u = 2.45 - 7.00a
u = 2.45 - 7.00(-0.93)
u = 2.45 + 6.51
u ≈ 8.96 m/s
Thus, the truck's original speed was approximately 8.96 m/s, and the acceleration was approximately -0.93 m/s^2.