7. On a user's manual of a motorbike you see a specification labeled as "the bike can accelerate to 100km/h from rest in 7 4seconds A) what can you say about the maximum average acceleration of the bike?
B) How far does it travel in this time?
8. The angle bbetween vector A and Vector B is 20⁰. If the magnitude of vector A is 6 unite and that B is 4 units.
What is the component of vectors along the direction of vector B.
9. A girl pushes a box placed in the floor with a force of 50N at an angle of 37⁰ below the horizontal. If the box moves towards the positive x direction. What are the vector components of the force?
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?
13. Given a position vector P=-4i + 6j What is the cosine of the angle between p and (a) the x-axis?. (b) the y-axis?. The cosine of the angle between a vector and the positive coordinate axis is called direction cosine of the vector.
14. A ball is thrown vertically upwards with a velocity of 10m/s from the balcony of a tall building. The balcony is 15m above the ground and gravitational acceleration is 10m/s². Find (a) the time required
for the ball to hit the ground, and (b) the velocity with which it hits the ground.
15. What vector must be added to vector C of 10km, east in order to give a resultant vector of?
A/15km,east.
B/ 15 km, West.
7. A) The maximum average acceleration of the bike can be determined by dividing the change in velocity by the time taken. In this case, the bike accelerates from rest to 100 km/h in 7.4 seconds, so the maximum average acceleration can be calculated as (100 km/h - 0 km/h) / 7.4 s = 13.51 km/h².
B) To find the distance traveled by the bike in this time, we can use the equation of motion: distance = initial velocity * time + 0.5 * acceleration * time². Since the initial velocity is 0 km/h and the acceleration is the maximum average acceleration calculated in part A, the distance can be calculated as: distance = 0 * 7.4 s + 0.5 * (13.51 km/h²) * (7.4 s)².
8. To find the component of vector A along the direction of vector B, we can use the formula: component of A along B = magnitude of A * cos(angle between A and B). In this case, the magnitude of vector A is 6 units and the angle between vector A and vector B is 20 degrees, so the component of vector A along the direction of vector B is: 6 units * cos(20 degrees).
9. The vector components of the force applied by the girl can be determined by resolving the force into its horizontal and vertical components. The horizontal component is 50 N * cos(37 degrees) towards the positive x direction, and the vertical component is 50 N * sin(37 degrees) in the negative y direction.
10. (a) To determine how far beyond the traffic signal the automobile will overtake the truck, we need to find the time it takes for the automobile to catch up with the truck. Since the truck is moving with a constant speed of 10 m/s and the automobile is accelerating at 5.0 m/s², we can use the formula: distance = initial velocity * time + 0.5 * acceleration * time². Setting the initial velocity of the automobile as 0 m/s, the acceleration as 5.0 m/s², and solving for time, we can then use the formula distance = speed * time to determine the distance.
(b) The velocity of the automobile at the instant it overtakes the truck can be found by using the formula: final velocity = initial velocity + acceleration * time.
11. A) The angle between two vectors can be found using the dot product formula: angle = arccos((A dot B) / (magnitude of A * magnitude of B)). In this case, the angle between X and Y can be calculated as arccos((4 * 6 + 1 * 3) / (sqrt(4²+1²) * sqrt(6²+3²))).
B) The component of vector X along the direction of vector Y can be calculated using the formula: component of X along Y = magnitude of X * cos(angle between X and Y). In this case, the magnitude of vector X is given as 4i + j, and the angle between X and Y can be found using the dot product formula mentioned in part A.
12. To determine if the train will stop in time, we need to find the time it takes for the train to come to a stop. Using the formula: final velocity² = initial velocity² + 2 * acceleration * distance, we can solve for time. If the calculated time is less than the time it takes for the train to reach the end of the platform, which can be found by dividing the platform length by the initial velocity, then the train will stop in time.
13. (a) The cosine of the angle between vector P and the x-axis can be found using the formula: cosine of the angle = P dot i / (magnitude of P * magnitude of i). In this case, the magnitude of vector P is given as -4i + 6j, and the magnitude of i is 1 unit.
(b) The cosine of the angle between vector P and the y-axis can be found using the formula: cosine of the angle = P dot j / (magnitude of P * magnitude of j). In this case, the magnitude of vector P is given as -4i + 6j, and the magnitude of j is 1 unit.
14. (a) The time required for the ball to hit the ground can be found using the kinematic equation: final velocity = initial velocity + acceleration * time. In this case, the initial velocity is 10 m/s, the final velocity is 0 m/s (since it hits the ground), and the acceleration is -10 m/s² (due to gravity). Solving for time will give the required answer.
(b) The velocity with which the ball hits the ground can be found using the formula: final velocity = initial velocity + acceleration * time. In this case, the initial velocity is 10 m/s, the acceleration is -10 m/s², and the time can be found in part (a). Solving for final velocity will give the required answer.
15. A) To find the vector that must be added to vector C of 10 km, east in order to give a resultant vector of 15 km, east, we can subtract vector C from the resultant vector. Therefore, the required vector is 15 km, east - 10 km, east.
B) To find the vector that must be added to vector C of 10 km, east in order to give a resultant vector of 15 km, west, we can subtract vector C from the resultant vector. Therefore, the required vector is 15 km, west - 10 km, east.