Concept explainers
(a)
The distance between the nose of car and the south edge when the car stops.
(a)
Answer to Problem 81CP
The distance between the nose of car and the south edge when the car stops is
Explanation of Solution
Write the expression for the final position of the nose of the car.
Here,
The initial position of the car is zero. Put
Write the expression for the final velocity of the car.
Here,
Rearrange expression (III) to find
Conclusion:
Substitute
Since the blue car stops at the intersection, final velocity is zero.
Substitute
Therefore, the distance between the nose of car and the south edge when the car stops is
(b)
The time interval in which the car is in the boundaries of intersection.
(b)
Answer to Problem 81CP
The time in which the car is in the boundaries of intersection is
Explanation of Solution
The time for which the car is in the intersection is the time between the entering of nose and exiting of tail from the intersection. Thus the total distance travelled by the car between the intersections is equal to the sum of length of car and length of the intersection path. Thus the change in position of nose of the car is equal to
Write the expression for the final position of car at time
The car starts from origin, so
Expression (V) is a quadratic equation.
Write the general expression for a quadratic equation in terms of
Here,
Write the expression to find the solution for quadratic equation (VI).
Conclusion:
Substitute
Compare the above quadratic expression with (VI) to obtain the values of constants.
Substitute
Thus the time values are
Therefore, the time in which the car is in the boundaries of intersection is
(c)
The minimum distance from the near edge of intersection where the red car can start its motion after the complete leaving of blue car.
(c)
Answer to Problem 81CP
The minimum distance from the near edge of intersection where the red car can starts its motion after the complete leaving of blue car is
Explanation of Solution
The nose of the blue car enters the intersection at
Again use expression (I) to find the distance between near edge and nose of car.
Conclusion:
Substitute
Therefore, the minimum distance from the near edge of intersection where the red car can starts its motion after the complete leaving of blue car is
(d)
The speed of red car when it enters the intersection.
(d)
Answer to Problem 81CP
The speed of red car when it enters the intersection is
Explanation of Solution
Write the expression to find the velocity of red car.
Conclusion:
Substitute
Therefore, the speed of red car when it enters the intersection is
Want to see more full solutions like this?
Chapter 2 Solutions
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
- A particle moves along the x axis according to the equation x = 2.00 + 3.00t 1.00t2, where x is in meters and t is in seconds. At t = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.arrow_forwardAs some molten metal splashes, one droplet flies off to the east with initial velocity vi at angle i above the horizontal, and another droplet flies off to the west with the same speed at the same angle above the horizontal as shown in Figure P4.40. In terms of vi and i, find the distance between the two droplets as a function of time. Figure P4.40arrow_forwardA rail gun shoots a projectile in a straight line with acceleration aa and in time tt as defined by the equation a(t)=3−2t. If the projectile from the launcher reaches a velocity of 10 at t=1 and if s(t) is the distance of the projectile from the rail gun at time t, find s(5)−s(1)arrow_forward
- Chinook salmon can cover more distance in less time by periodically making jumps out of the water. Suppose a salmon swimming in still water jumps out of the water with velocity 6.03 m/s at 40.6° above the horizontal, re-enters the water a distance L upstream, and then swims the same distance L underwater in a straight, horizontal line with velocity 2.92 m/s before jumping out again. Where L=3.67 m; t1= 0.801 sec this is a time when fish are over the water; t2= 1.256 sec; Ttotal =2.057 sec; the average horizontal velocity is 3.57 m/s. Consider the interval of time necessary to travel 2L. How is this reduced by the combination of jumping and swimming compared with just swimming at the constant speed of 2.92 m/s? Express the reduction as a percentage.arrow_forwardAt t = 0, one toy car is set rolling on a straight track with initial position 13.0 cm, initial velocity -2.8 cm/s, and constant acceleration 2.30 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 9.5 cm, initial velocity 6.00 cm/s, and constant zero acceleration. (a) At what time, if any, do the two cars have equal speeds?arrow_forwardA jet plane comes in for a landing with a speed of 100m/s, and its acceleration can have a maximum magnitude of 5.00m/s2 as it comes to rest from the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest?arrow_forward
- On a spacecraft two engines fire for a time of 532 s. One gives the craft an acceleration in the x direction of ax = 4.92 m/s2, while the other produces an acceleration in the y direction of ay = 7.16 m/s2. At the end of the firing period, the craft has velocity components of vx = 3743 m/s and vy = 4808 m/s. Calculate the magnitude of the initial velocity.arrow_forwardA particle performs a one-dimensional motion with the position given by the time equation x(t) = 1,3t4 - 2,0t3, where x is given in meters and t is given in seconds. At time t=0, the particle starts its motion at the origin x=0. At what instant (in seconds) does the particle reverse its direction of motion?arrow_forwardA body moving in the positive x-direction passes the origin at time t = 0. Between t = 0 and t = 1 second, the body has a constant speed of 14 (m/s). At t = 1 second, the body is given a constant acceleration of 5 (m/s2) in the negative x-direction. The position x of the body in meters at t = 9 seconds isarrow_forward
- A jet plane comes in for a landing with a speed of 100 m/s, and its acceleration can have a maximum magnitude of 5.00 m/s^2 as it comes to rest, from the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest?arrow_forwardAn object is moving along the x-axis. At t = 0 it is at x = 0. Its x-component of velocity vx as a function of time is given by vx(t)=αt−βt3, where α=α= 7.6 m/s2 and β= 4.4 m/s4. At what nonzero time tt is the object again at xx = 0?arrow_forwardA particle is moving in a straight line with its position given by x=(8.5m/s^2)t^2 + 6.0m. Calculate. a) The instantaneous velocity at t1=3.00s and t2=5.00s. b) It's average acceleration during the interval from t1=3.00s to t2=5.00s. c) Its instantaneous acceleration as a function of time.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning