A car is travelling at 80km/h in east direction , a train is travelling at 50 m/s in north direction then what is the combined velocity of car and train?
They should be asking for what is the velocity of the car with respect to the train, or vice versa. The "combined velocity" is the vector sum. The relative velocity is the vector difference.
In this case, they both have the same magnitude. Use the Pythagorean theorem, since the separate velocities are at right angles.
. A 2000 kg car traveling at 8.0 m/s has a rear-end collision with a 1500 kg car stopped at a traffic light. The cars lock bumpers and skid off together.
a. Which of the ranges below is correct for the resulting velocity?
To find the combined velocity of the car and train, we need to consider both the magnitude (speed) and direction of their velocities.
First, let's convert the car's velocity from km/h to m/s. To do this, we need to divide the speed in km/h by 3.6 since there are 3.6 seconds in an hour.
80 km/h ÷ 3.6 = 22.22 m/s (rounded to two decimal places)
Now we have the speed of the car as 22.22 m/s in the east direction.
To calculate the combined velocity, we need to add the velocity vectors. Since they are at right angles to each other (east and north direction), we can use the Pythagorean theorem to find the magnitude of the combined velocity.
Magnitude of the combined velocity = √(velocity of car)^2 + (velocity of train)^2
Magnitude of the combined velocity = √(22.22 m/s)^2 + (50 m/s)^2
Magnitude of the combined velocity = √(492.49 m^2/s^2 + 2500 m^2/s^2)
Magnitude of the combined velocity = √(2992.49 m^2/s^2)
Magnitude of the combined velocity ≈ 54.69 m/s (rounded to two decimal places)
Since both the car and the train are traveling in different directions, it's important to note that the combined velocity will have both magnitude and direction. To specify the direction, we can use trigonometry to find the angle between the combined velocity vector and the east direction.
To find the angle, we can use the inverse tangent function:
Angle = arctan((velocity of train)/(velocity of car))
Angle = arctan(50 m/s / 22.22 m/s)
Angle ≈ 66.80° (rounded to two decimal places)
So, the combined velocity of the car and the train is approximately 54.69 m/s in a direction approximately 66.80° north of east.