A car and a truck starts from rest at the same instant, with the car initially at some distant behind the truck. The truck has a constant acceleration of 2.1 m/s^2, and the car at 3.4 m/s^2. The car overtakes the ruck after it has moved 40.0 m.
a) how much time does it take the car to overtake the truck?
b)how far does the car behind the truck initially?
c)what is the speed of each when they abreast?
d)on a single graph, sketch the position of each vehicle as a function of time. take x=0 at the initial location of the truck.
To solve this problem, we can use kinematic equations, specifically the equation of motion:
x = x0 + v0*t + (1/2)*a*t^2
where:
x = final position
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
Let's solve the problem step by step:
a) To determine how much time it takes for the car to overtake the truck, we need to calculate the time when their positions are equal (abreast).
Let's assume the distance traveled by the car is x_car and the distance traveled by the truck is x_truck. When the car overtakes the truck, their positions are the same, so we have:
x_car = x_truck
Using the equation of motion for the car:
x_car = 0 + 0 + (1/2) * 3.4 * t^2 = 1.7 * t^2
Using the equation of motion for the truck:
x_truck = 0 + 0 + (1/2) * 2.1 * t^2 = 1.05 * t^2
Setting these two expressions equal:
1.7 * t^2 = 1.05 * t^2
Simplifying:
0.65 * t^2 = 0
Since the car overtakes the truck after it has moved 40.0 m, we can set up another equation:
x_car = 1.7 * t^2 = 40.0
Solving for t:
1.7 * t^2 = 40.0
t^2 = 40.0 / 1.7
t^2 = 23.529
t ≈ √23.529
t ≈ 4.85 seconds
Therefore, it takes approximately 4.85 seconds for the car to overtake the truck.
b) To determine how far the car is initially behind the truck, we need to calculate the distance traveled by the truck when the car overtakes it.
Using the equation for the truck:
x_truck = 0 + 0 + (1/2) * 2.1 * t^2 = 1.05 * t^2
Substituting the value of t:
x_truck = 1.05 * (4.85)^2
x_truck ≈ 23.0 meters
Therefore, the car is initially 23.0 meters behind the truck.
c) To find the speed of each vehicle when they are abreast, we need to calculate their respective velocities at that time.
Using the equation of motion for the car:
v_car = v0_car + a_car * t
v_car = 0 + 3.4 * 4.85
v_car ≈ 16.49 m/s
Using the equation of motion for the truck:
v_truck = v0_truck + a_truck * t
v_truck = 0 + 2.1 * 4.85
v_truck ≈ 10.17 m/s
Therefore, when they are abreast, the car's speed is approximately 16.49 m/s, and the truck's speed is approximately 10.17 m/s.
d) To sketch the position of each vehicle as a function of time on a single graph, we will place time on the x-axis and position on the y-axis.
Assuming the truck starts at position 0, the position of the truck as a function of time can be represented by the equation:
x_truck = 0 + 0 + (1/2) * 2.1 * t^2 = 1.05 * t^2
The position of the car as a function of time can be represented by the equation:
x_car = 0 + 0 + (1/2) * 3.4 * t^2 = 1.7 * t^2
Plotting these two equations on the same graph, you will see that initially, the car is behind the truck, and then it overtakes the truck at some time (approximately 4.85 seconds), becoming abreast.
Note: The graph will have time (t) on the x-axis and position (x) on the y-axis, with the positions of the truck and car being represented by the equations provided above.
a) Truck --> 40 = x0 + v0t + (1/2)at^2
40 = (1/2)at^2
t = sqrt(80/2.1)
b) Car x = x0 + v0t + (1/2)at^2
x0 = (1/2)(3.4)t^2 - 40
c) Car v = v0 + at --> v = 3.4t
Truck v = v0 + at --> v = 2.1t
d) Plug in values and prosper.