10. At the instant the traffic light turns green, an automobile starts moving with a constant acceleration of 5.0 m/s². At the same instant a truck, travelling with a constant speed of 10 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be at that instant?
11. Given two vectors X and Y such that X=4i+j and Y=6i+3j
A.What is the angle b/n the two vectors. B/ Find the component of X along the direction of Y.
12. The driver of a train travelling at 40 m/s applies the brakes as the train enters a station. The train slows down
at a rate of 2m/s. The platform is 400 m long. Will the train stop in time?
10. To solve this problem, we can use the kinematic equation for motion with constant acceleration:
d = vit + (1/2)at^2
where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time.
(a) Let's assume that the time it takes for the automobile to overtake the truck is t. During this time, the truck also travels a distance given by d = vit (since it has a constant velocity). Therefore, the distance traveled by the automobile is the sum of the distances traveled by the truck and itself, which can be represented by the equation:
d = vit + (1/2)at^2
However, since the initial velocity of the automobile is 0 (it starts from rest when the traffic light turns green), the equation simplifies to:
d = (1/2)at^2
Plugging in the values for the acceleration (a = 5.0 m/s^2), we can solve for t:
t^2 = 2d/a
Since the truck travels at a constant speed of 10 m/s, the time taken by the truck to travel a distance d is t = d/10.
Setting these two equations equal to each other and solving for d:
2d/a = d/10
20d = ad
d = 20a/2
d = 10a
So, the automobile overtakes the truck at a distance of 10 times the acceleration of the automobile.
(b) To find the velocity of the automobile at the moment it overtakes the truck, we can use the equation:
v = vi + at
Since the initial velocity of the automobile is 0, the equation simplifies to:
v = at
Plugging in the values for the acceleration (a = 5.0 m/s^2), we can find the velocity:
v = 5.0 m/s^2 * t
Using the value of t from part (a), we can calculate the velocity:
v = 5.0 m/s^2 * (10a/2) = 25a m/s
So, the velocity of the automobile at the moment it overtakes the truck is 25 times the acceleration of the automobile.
11.
(a) To find the angle between the two vectors, we can use the dot product formula:
dot(X, Y) = |X| * |Y| * cos(theta)
where dot(X, Y) represents the dot product of X and Y, |X| and |Y| represent the magnitudes of X and Y, and theta represents the angle between X and Y.
The magnitude of X is |X| = sqrt(4^2 + 1^2) = sqrt(17).
The magnitude of Y is |Y| = sqrt(6^2 + 3^2) = sqrt(45).
The dot product of X and Y is dot(X, Y) = 4 * 6 + 1 * 3 = 27.
Plugging in these values into the dot product formula:
27 = sqrt(17) * sqrt(45) * cos(theta)
Solving for cos(theta):
cos(theta) = 27 / (sqrt(17) * sqrt(45))
Taking the inverse cosine of both sides to find theta:
theta = acos(27 / (sqrt(17) * sqrt(45)))
Using a calculator, we find that theta is approximately 26.6 degrees.
So, the angle between the two vectors is approximately 26.6 degrees.
(b) To find the component of X along the direction of Y, we can use the formula:
component of X along Y = |X| * cos(theta)
where |X| is the magnitude of X and theta is the angle between X and Y.
Using the values we previously calculated:
component of X along Y = sqrt(17) * cos(theta)
Plugging in the value of theta:
component of X along Y = sqrt(17) * cos(26.6)
Using a calculator, we find that the component of X along the direction of Y is approximately 11.82.
So, the component of X along the direction of Y is approximately 11.82.
12. To determine if the train will stop in time, we need to find the distance it takes for the train to stop given its initial speed and the deceleration. We can use the following equation:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s since the train stops), u is the initial velocity (40 m/s), a is the acceleration (-2 m/s^2), and s is the distance traveled.
Simplifying the equation:
0 = (40 m/s)^2 + 2(-2 m/s^2)(s)
0 = 1600 m/s^2 - 4 m/s^2 * s
4 m/s^2 * s = 1600 m/s^2
s = 400 m
The distance traveled by the train is 400 m, which is equal to the length of the platform. Therefore, the train will stop in time before reaching the end of the platform.
So, the train will stop in time before reaching the end of the platform.