A truck covers 40.0 m in 8.25 s while uniformly slowing down to a final velocity of 3.35 m/s.
(a) Find the truck's original speed.
m/s
(b) Find its acceleration.
m/s2
To find the truck's original speed and its acceleration, we can use the equations of motion.
Let's start with equation (1):
\[v = u + at\]
Where:
v = final velocity (3.35 m/s)
u = initial velocity (to be found)
a = acceleration (to be found)
t = time taken (8.25 s)
(a) Find the truck's original speed (u):
Rearranging equation (1), we get:
\[u = v - at\]
Substituting the given values, we get:
\[u = 3.35 m/s - a(8.25 s)\] (Equation 2)
Now, let's consider the equation for distance covered (displacement):
\[s = ut + \frac{1}{2}a t^2\]
Where:
s = distance covered (40.0 m)
u = initial velocity (to be found)
a = acceleration (to be found)
t = time taken (8.25 s)
Rearranging the equation, we get:
\[a = \frac{2(s - ut)}{t^2}\] (Equation 3)
Since we know the final velocity is 3.35 m/s, we can substitute it into equation (3) to get:
\[a = \frac{2(s - 3.35 t)}{t^2}\] (Equation 4)
We can now solve equations (2) and (4) simultaneously to find the values of u and a.
Let's substitute equation (2) into equation (4):
\[a = \frac{2(s - (3.35 - a(8.25)) \cdot (8.25^2)}{8.25^2}\]
Now, solve this equation for a:
\[a = \frac{2(40 - 27.6375 + 8.25a)}{68.0625}\]
Multiply both sides of the equation by 68.0625:
\[68.0625a = 2(40 - 27.6375 + 8.25a)\]
\[68.0625a = 2(40 - 27.6375) + 16.5a\]
\[68.0625a - 16.5a = 2(12.3625)\]
\[51.5625a = 24.725\]
Divide both sides of the equation by 51.5625:
\[a = \frac{24.725}{51.5625}\]
\[a \approx 0.479 m/s^2\]
Now, substitute the value of a into equation (2) to solve for u:
\[u = 3.35 - 0.479(8.25)\]
\[u = 3.35 - 3.954\]
\[u \approx -0.604 m/s\] (Note: The negative sign indicates that the truck was moving in the opposite direction.)
Therefore,
(a) The truck's original speed is approximately -0.604 m/s.
(b) The truck's acceleration is approximately 0.479 m/s^2.