Will Car A Collide with Car B After Deceleration?

In summary, two cars, A and B, both 5m in length, are traveling with constant velocity of 20 m/s along a straight level road. When they reach point P, each car decelerates at a rate of 4m/s^2. The crucial point is that car A starts decelerating one second after car B. The collision occurs when the front of car A is at the same point as the back of car B, and this happens after 5 seconds at a distance of 20m from point P. The motion of both cars can be shown on the same speed-time graph, with car B starting at 20m/s and car A starting at 0m/s.
  • #1
mcintyre_ie
66
0
Hey, id greatly appreciate any urgent help you can give me with this problem

Two cars A and B, each 5m in length, travel with constant velocity 20 m/s along a straight level road. The front of car A is 15m directly behind the rear of car B. Immediately on reaching a point P, each car decelerates at 4m/s^2.
1) Show that A collides with B
2) At what distance from P does the collision occur?
3) Show the motion of both cars on the same speed-time graph.


Thanks in advance
 
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  • #2
Please! I am seriously desperate here!
 
  • #3
I know! You also posted this on the "mathematics" forum.

A crucial point here is that since each car begins decelerating when it "reaches point P", car A does not begin decelerating until AFTER car B does. In fact, since the front of car A is 20 m behind the front of car B (I assume they "reach point P" when the front of the car does) and car A is moving at 20 m/s, car A will start decelerating exactly 1 second after car B does.
With constant deceleration of 4 m/s2 and taking t= 0 when the front of car B passes point P, vB(t) (the speed of car B at time t) is 20- 4t and so xB(t) (the distance from point P to the front of car B at time t) is 20t- 2t2.

A has the same initial speed and the same deceleration but starts a second later: vA= 20- 4(t-1) and
xA(t)= 20(t-1)-2(t-1)2 (notice the clever use of (t-1) as the variable- I could have said the anti-derivative of 20 is 20t but then I would have to add -20 to make xA= 0 when t= 1 rather than 0.)

The two cars will collide when the FRONT of car A is at the same point as the BACK of car B. Since xB and xA are the distance from P to the fronts of the cars, the collision will happen when xB- xA= 5. Set up that equation and solve for t. Once you find that, substitute t into the equations for xB and xA to find the distances.
 

FAQ: Will Car A Collide with Car B After Deceleration?

What is accelerated linear motion?

Accelerated linear motion is the motion of an object where its velocity is changing at a constant rate, resulting in a change in its position over time. This type of motion is characterized by an increase or decrease in speed, as well as a change in direction.

What factors affect accelerated linear motion?

The factors that affect accelerated linear motion include the initial velocity of the object, the rate of change in velocity (acceleration), and the time interval over which the motion occurs. Other external factors such as friction, air resistance, and gravity can also impact the motion of an object.

How is accelerated linear motion different from uniform linear motion?

Uniform linear motion is when an object moves at a constant speed in a straight line, while accelerated linear motion involves a change in velocity over time. In uniform linear motion, the acceleration is zero, while in accelerated linear motion, the acceleration is a non-zero constant value.

What are some real-life examples of accelerated linear motion?

Some real-life examples of accelerated linear motion include a car accelerating from a stop sign, a rollercoaster going down a steep drop, a skydiver falling towards the ground, and a rocket launching into space. These are all instances where the object's velocity is changing at a constant rate, resulting in accelerated linear motion.

How is accelerated linear motion represented mathematically?

Accelerated linear motion can be represented mathematically using the equations of motion, such as v = u + at for finding the final velocity, s = ut + 1/2at^2 for finding the displacement, and v^2 = u^2 + 2as for finding the final velocity without time. These equations involve the variables of initial velocity (u), final velocity (v), acceleration (a), and time (t), and can be used to calculate various quantities related to accelerated linear motion.

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