On its way to Peace River, Alberta, a car enters a passing zone on the highway. The car is 25.0 m behind a truck and both vehicles are travelling at 41.0 m/s. The car accelerates uniformly to overtake the truck and after the car has travelled 280m, it is 25.0 m ahead of the truck, which has continued to travel at 41 m/s. How much time elapses between when the car begins to accelerate and when the car is 25.0 m ahead of the truck?

Question

On its way to Peace River, Alberta, a car enters a passing zone on the highway. The car is 25.0 m behind a truck and both vehicles are travelling at 41.0 m/s. The car accelerates uniformly to overtake the truck and after the car has travelled 280m, it is 25.0 m ahead of the truck, which has continued to travel at 41 m/s. How much time elapses between when the car begins to accelerate and when the car is 25.0 m ahead of the truck?

Answer 1

The truck travels

X1 = 41 t meters
during the interval t that it takes to go from 25 m behind to 25 m ahead. The car travels
H2 = 41 t + (1/2) a t^2 = 41 t + 50 = 280

41 t = 230
t = 5.61 s

Answer 2

To solve this problem, we need to determine the time it takes for the car to accelerate and travel a distance of 280m.

First, let's calculate the initial velocity of the car. Since both vehicles are traveling at 41.0 m/s initially and the car starts from behind the truck, its initial velocity relative to the truck is 0 m/s.

Next, we need to find the acceleration of the car. We can use the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

In this case, the final velocity of the car is 41.0 m/s (the same as the truck's velocity) and the distance traveled is 280m. Plugging in the values into the equation, we have:

41.0^2 = 0^2 + 2a(280)

1681 = 560a

Now, let's solve for the acceleration:

a = 1681 / 560
a = 3.00 m/s^2

Now that we know the acceleration, we can determine the time it takes for the car to travel the 280m distance and be 25.0m ahead of the truck. We can use the kinematic equation:

s = ut + 0.5at^2

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values, we have:

280 = 0*t + 0.5*3.00*t^2

280 = 1.5t^2

Dividing both sides by 1.5:

t^2 = 280 / 1.5
t^2 = 186.67

Taking the square root of both sides:

t = sqrt(186.67)
t = 13.66 s

Therefore, it takes approximately 13.66 seconds for the car to accelerate and travel a distance of 280m, being 25.0m ahead of the truck.