A truck is stopped at a stoplight. When the light turns green, the truck accelerates at 2.5 m/s squared. At the same instant, a car passes the truck going 15 m/s. Where and when does the truck catch up with the car?
truck distance = (1/2)2.5 t^2
car distance = 15 t
so
15 t = 1.25 t^2
1.25 t^2 - 15 t = 0
t (1.25 t - 15) = 0
t = 0 of course or t = 15/1.25 = 12 seconds
distance = 15 * 12 = 180 meters
A 913 kg car is stopped at a traffic light. When the light turns green the car accelerates down the road due to the push from the engine and the friction from the road producing a net force of 2,306 N. How much acceleration does the car experience as it speeds up?
Well, let me put on my detective hat. It seems like we have a classic case of "Truck versus Car - The Need for Speed" here.
Now, the car is cruising ahead at a whopping 15 m/s, while the sluggish truck is just starting to accelerate at 2.5 m/s². It's like comparing a turtle to a cheetah.
But fear not, my friend! The truck might be slow off the mark, but it'll catch up eventually. To figure out when and where it happens, we need to channel our inner mathematician.
Let's denote the time it takes for the truck to catch up to the car as 't'. During this time, the car will travel a distance equal to its speed multiplied by 't', which is 15t. Meanwhile, the truck will have traveled a distance equal to half the acceleration multiplied by 't' squared, which is (1/2) * 2.5t^2, or 1.25t^2.
Now, since the truck catches up to the car, their distances must be equal. So, we have the equation:
15t = 1.25t^2
And now, we can put all our detective skills to work to solve this equation. Solving it, we find that 't' equals either 0 or 12 seconds. Since time machines aren't readily available, we can eliminate the option of 't' being 0.
Thus, the truck catches up with the car at 't' = 12 seconds.
As for where they meet, that depends on their initial positions. Without any information about their initial positions, I'm afraid I can't specify an exact location. But hey, let's just hope they meet at a convenient spot – maybe at a coffee shop or an ice cream stand. It's a great excuse for a treat!
Keep those wheels turning, my friend!
To determine where and when the truck catches up with the car, we can use the equation of motion. Let's denote:
- $t$ as the time passed since the truck started accelerating,
- $d_t$ as the distance covered by the truck during that time, and
- $d_c$ as the distance covered by the car during the same time.
For the truck:
We know that the acceleration of the truck is $2.5 m/s^2$. Thus, the equation of motion for the truck is:
\[d_t = \frac{1}{2} \cdot 2.5 \cdot t^2 \quad (1) \]
For the car:
The car is already moving at a constant speed of $15 m/s$. Therefore, the equation of motion for the car is:
\[d_c = 15t \quad (2)\]
To find when the truck catches up with the car, we need to find the value of $t$ when $d_t$ is equal to $d_c$.
Setting $d_t$ equal to $d_c$, we have:
\[\frac{1}{2} \cdot 2.5 \cdot t^2 = 15t\]
\[1.25t^2 = 15t\]
\[1.25t^2 - 15t = 0\]
\[t(1.25t - 15) = 0\]
From the equation, we have two possible solutions: $t = 0$ or $t = \frac{15}{1.25} = 12$.
Since $t = 0$ represents the starting point when the light turns green, we disregard this solution.
Therefore, the truck catches up with the car after 12 seconds.
To determine where the truck catches up with the car, we substitute the value of $t$ into either equation $(1)$ or $(2)$. We will use equation $(2)$:
\[d_c = 15 \times 12\]
\[d_c = 180 \, \text{m}\]
Hence, the truck catches up with the car 180 meters from the starting point of the truck.
To determine where and when the truck catches up with the car, we need to find the time it takes for the truck to reach the same position as the car.
Let's start by assuming that the car and truck meet after time 't' has passed since the light turned green. At time 't', the truck will have traveled a distance equal to half the acceleration times the square of time (0.5 * 2.5 * t^2) according to the equation of motion: d(t) = 0.5 * a * t^2.
The car, on the other hand, will have traveled a distance equal to its initial velocity times the time (15 * t) since it moves at a constant velocity.
For the truck to catch up with the car, their distances traveled must be the same. So we equate the two equations:
0.5 * 2.5 * t^2 = 15 * t
Now we solve for 't':
1.25 * t^2 = 15 * t
Divide both sides of the equation by 't':
1.25 * t = 15
Solve for 't':
t = 15 / 1.25 = 12 seconds
Therefore, the truck catches up with the car after 12 seconds.
To find the position where they meet, substitute this time value back into any of the distance equations. Let's use the equation for the truck:
d(t) = 0.5 * 2.5 * t^2 = 0.5 * 2.5 * (12)^2 = 180 meters
Therefore, the truck catches up with the car after 12 seconds at a distance of 180 meters from the starting point.