A pickup truck is stopped at a red light. When the light turns green and the pickup just starts to accelerate uniformly at 5.0 m/s2, a motorcycle travelling at a constant 18 m/s passes the pickup in the adjacent lane. How far will the pickup truck travel before it overtakes the motorcycle?
I'd like it if you provide me the equations needed for the problem, it's been years that I've touched any mathematics so this is really confusing to me. (So far I've learned kinematics and velocity-time/position-time graph equations) It's not necessarily time-limited but I just want to understand how it works. Thanks
To solve this problem, we can use the equations of motion in kinematics. Here are the equations that will be helpful:
1. v = u + at: This equation relates the final velocity (v) to the initial velocity (u), acceleration (a), and time (t).
2. s = ut + (1/2)at^2: This equation relates the distance traveled (s) to the initial velocity (u), time (t), and acceleration (a).
3. v^2 = u^2 + 2as: This equation relates the final velocity (v) to the initial velocity (u), distance traveled (s), and acceleration (a).
Let's define the variables for this problem:
u1 = initial velocity of the pickup truck (which is 0 m/s since it is stopped)
a1 = acceleration of the pickup truck (5.0 m/s^2)
u2 = initial velocity of the motorcycle (18 m/s)
a2 = acceleration of the motorcycle (0 m/s^2 since it is traveling at a constant speed)
s = distance traveled by the pickup truck before overtaking the motorcycle
t = time taken by the pickup truck to overtake the motorcycle
As the motorcycle is already passing the pickup truck when the light turns green, we can assume that the pickup truck starts from rest. Therefore, using equation 1, we can find the time taken by the pickup truck to reach the same velocity as the motorcycle:
v1 = u1 + a1t
18 = 0 + 5.0t
t = 18/5.0
t = 3.6 seconds
Now we can use equation 2 to find the distance traveled by the pickup truck in this time:
s = u1t + (1/2)a1t^2
s = 0 * 3.6 + (1/2) * 5.0 * (3.6)^2
s = 0 + (1/2) * 5.0 * 12.96
s = 32.4 meters
Therefore, the pickup truck will travel a distance of 32.4 meters before overtaking the motorcycle.
To solve this problem, we need to use the equations of motion. The two equations we need are:
1. v = u + at
This equation relates the final velocity (v), initial velocity (u), acceleration (a), and time (t).
2. s = ut + (1/2)at^2
This equation relates the displacement (s), initial velocity (u), time (t), and acceleration (a).
Let's denote the initial velocity of the pickup truck as u1, and the initial velocity of the motorcycle as u2. We are given:
u1 = 0 m/s (as the pickup is initially stopped)
u2 = 18 m/s (constant velocity of the motorcycle)
a = 5.0 m/s^2 (acceleration of the pickup)
We need to find the time it takes for the pickup truck to overtake the motorcycle. To do this, we can set the final displacement of the pickup truck (s1) equal to the displacement of the motorcycle (s2) when the pickup truck catches up to it.
So, in equation form:
s1 = s2
Using equation 2, we can write:
u1t + (1/2)a1t^2 = u2t
Since u1 = 0, the equation simplifies to:
(1/2)a1t^2 = u2t
Rearranging the equation:
(1/2)a1t = u2
Simplifying further:
(1/2)(5.0)t = 18
2.5t = 18
t = 18 / 2.5
t = 7.2 seconds
Now, we can find the distance traveled by the pickup truck during this time using equation 2:
s1 = u1t + (1/2)a1t^2
Substituting the values we have:
s1 = 0 * 7.2 + (1/2)(5.0)(7.2)^2
Simplifying:
s1 = 0 + (1/2)(5.0)(51.84)
s1 = 129.6 meters
Therefore, the pickup truck will travel a distance of 129.6 meters before it overtakes the motorcycle.
To solve this problem, we can use the kinematic equations of motion. In particular, we will use the equation:
\[d = v_0 t + \frac{1}{2} a t^2\]
where:
- \(d\) is the distance traveled by the pickup truck
- \(v_0\) is the initial velocity of the pickup truck (which is 0 in this case since it starts from rest)
- \(a\) is the acceleration of the pickup truck (given as 5.0 m/s²)
- \(t\) is the time taken by the pickup truck to overtake the motorcycle.
We know that the motorcycle is traveling at a constant velocity of 18 m/s. Since the pickup truck needs to overtake the motorcycle, it needs to catch up to it and match its velocity. Therefore, we need to find the time it takes for the pickup truck to catch up to the motorcycle.
Let's assume that the time it takes for the pickup truck to overtake the motorcycle is \(t\) seconds. During this time, the motorcycle will have traveled a distance equal to \(18t\) meters.
Now, let's calculate the distance traveled by the pickup truck during the same time \(t\). We can use the kinematic equation mentioned earlier, with \(v_0 = 0\) m/s, \(a = 5.0\) m/s², and \(d\) as the unknown distance:
\[d = 0 \cdot t + \frac{1}{2} \cdot 5.0 \cdot t^2\]
Simplifying the equation gives us:
\[d = \frac{5}{2} t^2\]
Now, we need to find the value of \(t\) when the distance traveled by the pickup truck is equal to the distance covered by the motorcycle. So we equate the two distances:
\[\frac{5}{2} t^2 = 18t\]
Rearranging the equation:
\[\frac{5}{2} t^2 - 18t = 0\]
Now, we can solve this quadratic equation for \(t\). Factoring out \(t\), we get:
\[t \left(\frac{5}{2} t - 18\right) = 0\]
This equation has two solutions: \(t = 0\) (which is not meaningful as the motorcycle is already ahead of the pickup truck) and:
\[\frac{5}{2} t - 18 = 0\]
Solving for \(t\), we find:
\[\frac{5}{2} t = 18\]
\[t = \frac{36}{5}\] seconds
Using this value of \(t\), we can find the distance traveled by the pickup truck:
\[d = \frac{5}{2} \cdot \left(\frac{36}{5}\right)^2\] meters
Evaluating the expression gives us the desired distance traveled by the pickup truck.
I hope this explanation helps. Let me know if you have any further questions!