At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.10m/s2 . At the same instant, a truck, traveling with a constant speed of 15.1m/s , overtakes and passes the automobile.

Question

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.10m/s2 . At the same instant, a truck, traveling with a constant speed of 15.1m/s , overtakes and passes the automobile.

How far beyond its starting point does the automobile overtake the truck?

How fast is the automobile traveling when it overtakes the truck?

could someone show me the formulas used for these I'm a little stuck

Answer 1

To solve this problem, we can use the equations of motion for uniformly accelerating objects. The key equations we will use are:

1) v = u + at
2) s = ut + 0.5at^2

Where:
- v is the final velocity
- u is the initial velocity
- a is the constant acceleration
- t is the time
- s is the distance

Let's start by calculating the time it takes for the automobile and the truck to meet.

For the automobile:
Initial velocity, u1 = 0 m/s (since it starts from rest)
Acceleration, a1 = 2.10 m/s^2

For the truck:
Velocity, v2 = 15.1 m/s
Acceleration, a2 = 0 m/s^2 (since the truck maintain a constant speed)

Using Equation 1 for both the automobile and the truck:

v1 = u1 + a1 * t
v2 = u2 + a2 * t

Since the automobile starts from rest:
v1 = a1 * t

We can solve for time, t:

t = v2 / a1
t = 15.1 m/s / 2.10 m/s^2
t = 7.19 s (rounded to two decimal places)

Now that we know the time it takes for the truck and the automobile to meet, we can calculate the distance the automobile traveled.

Using Equation 2 for the automobile:
s = u1 * t + 0.5 * a1 * t^2

s1 = 0 * 7.19 s + 0.5 * 2.10 m/s^2 * (7.19 s)^2
s1 = 0 + 0.5 * 2.10 m/s^2 * 51.72 s^2
s1 = 54.27 m (rounded to two decimal places)

Therefore, the automobile overtakes the truck by traveling 54.27 meters beyond its starting point.

To find the velocity (speed) of the automobile when it overtakes the truck, we can use Equation 1 for the automobile:

v1 = u1 + a1 * t

v1 = 0 m/s + 2.10 m/s^2 * 7.19 s
v1 = 15.12 m/s (rounded to two decimal places)

So, the velocity (speed) of the automobile when it overtakes the truck is 15.12 m/s.

Answer 2

To solve this problem, we can use the equations of motion for the two vehicles involved. Let's break down the problem into two parts: the automobile's motion up to the point it overtakes the truck, and the truck's motion during the same time period.

First, let's calculate the time it takes for the automobile to catch up to the truck. To find this, we can use the equation:

𝑡 = (∆𝑣/𝑎) + 𝑡₀

Where:
𝑡 = time taken
∆𝑣 = change in velocity
𝑎 = acceleration
𝑡₀ = initial time (which is 0 since they start at the same time)

In this case, the change in velocity is given by the difference between the truck's speed and the automobile's speed:

∆𝑣 = 𝑣𝑡𝑟𝑢𝑐𝑘 − 𝑣𝑎𝑢𝑡𝑜𝑚𝑜𝑏𝑖𝑙𝑒

Substituting the given values:
∆𝑣 = 15.1 m/s − 0 m/s = 15.1 m/s

Let's plug in the acceleration and solve for 𝑡:

𝑡 = (15.1 m/s)/(2.10 m/s²) = 7.19 seconds

Now that we have the value of 𝑡, we can calculate the distance traveled by the automobile up to the point it overtakes the truck. For this, we will use the equation:

𝑑 = 𝑣𝑎𝑢𝑡𝑜𝑚𝑜𝑏𝑖𝑙𝑒 × 𝑡 + (1/2) × 𝑎 × 𝑡²

Where:
𝑑 = distance traveled
𝑣𝑎𝑢𝑡𝑜𝑚𝑜𝑏𝑖𝑙𝑒 = initial velocity of the automobile
𝑡 = time taken
𝑎 = acceleration of the automobile

Given:
𝑣𝑎𝑢𝑡𝑜𝑚𝑜𝑏𝑖𝑙𝑒 = 0 m/s (since it starts from rest)
𝑡 = 7.19 s
𝑎 = 2.10 m/s²

Substituting these values into the equation:

𝑑 = (0 m/s) × (7.19 s) + (1/2) × (2.10 m/s²) × (7.19 s)²
= 51.16 m

Therefore, the automobile overtakes the truck at a distance of 51.16 meters beyond its starting point.

To calculate the speed of the automobile when it overtakes the truck, we need to find its velocity at that point. We can do this using the equation:

𝑣 = 𝑣₀ + 𝑎 × 𝑡

Where:
𝑣 = velocity of the automobile at the overtaking point
𝑣₀ = initial velocity of the automobile
𝑎 = acceleration of the automobile
𝑡 = time taken

Given:
𝑣₀ = 0 m/s (since it starts from rest)
𝑎 = 2.10 m/s²
𝑡 = 7.19 s

Substituting these values into the equation:

𝑣 = (0 m/s) + (2.10 m/s²) × (7.19 s)
= 15.10 m/s

Therefore, the automobile is traveling at a speed of 15.10 m/s when it overtakes the truck.