a physics student ‘borrows’ a sports car for a joy ride and discovers that it can accelerate at a rate of 4.90 m/s2. He decides to test the car by challenging Mr. Horn and his motorcycle. Both start from rest, but the student is so confident in his new ride that he gives Mr. Horn a 1.00 s head start. If Mr. Horn moves with a constant acceleration of 3.50 m/s2 and the student maintains his acceleration of 4.90 m/s2, find:
(a) the time it takes the student to overcome Mr. Horn.
(b) the distance he travels before he catches up with Mr. Horn.
(c) the speed of both vehicles at the instant the student overtakes Mr. Horn.
recall that
v = at
s = 1/2 at^2
the distances traveled at time t are
Horn: 3.50/2 t^2
student: 4.9/2 (t-1)^2
so now you can answer the questions
Mr. Horn travels 1.75 m during his 1 sec head start
... his velocity is 3.50 m/s when the student starts
(a) t is the time the student is moving ... Mr. Horn's time is 1 sec greater
... 4.90 t^2 = 3.50 t^2 + 3.50 t + 1.75 ... 1.40 t^2 - 3.50 t - 1.75 = 0
... t = 2.93 s
(b) d = 1/2 a t^2 = 1/2 * 4.90 * 2.93^2
(c) v = a t
... student ... v = 4.90 m/s^2 * 2.93 s
... Mr. Horn ... v = 3.50 m/s^2 * 3.93 s
oops ... didn't adjust the coefficients for Mr. Horn's travel
(a) 4.90 t^2 = 3.50 t^2 + 7.00 t + 3.50 ... 1.40 t^2 - 7.00 t - 3.50 = 0
... t = 5.46 s
plug the new time into (b) and (c)
sorry about that
To find the answers to these questions, we can use the equations of motion. Let's break down each question step by step.
(a) The time it takes the student to overcome Mr. Horn:
We can calculate the time it takes for each vehicle to reach the same distance. Since Mr. Horn has a 1.00 s head start, we need to find the time it takes for the student to catch up with him.
Let's denote the time it takes for the student to overtake Mr. Horn as 't'. Using the equation of motion for the student:
Distance covered by the student = initial velocity of the student * t + (1/2) * acceleration of the student * t^2
Since the student starts from rest, the initial velocity of the student is 0. Therefore, the equation simplifies to:
Distance covered by the student = (1/2) * acceleration of the student * t^2
Now, let's calculate the distance covered by Mr. Horn during the 1.00 s head start:
Distance covered by Mr. Horn = initial velocity of Mr. Horn * 1s + (1/2) * acceleration of Mr. Horn * (1s)^2
Since Mr. Horn starts from rest, his initial velocity is also 0, so the equation simplifies to:
Distance covered by Mr. Horn = (1/2) * acceleration of Mr. Horn * (1s)^2
Since the distance covered by both the student and Mr. Horn when the student overtakes Mr. Horn is the same, we can set up an equation:
(1/2) * acceleration of the student * t^2 = (1/2) * acceleration of Mr. Horn * (1s)^2
Now we can solve for 't':
(1/2) * 4.90 * t^2 = (1/2) * 3.50 * (1s)^2
Simplifying the equation, we have:
4.90 * t^2 = 3.50
Dividing both sides by 4.90 gives us:
t^2 = 3.50 / 4.90
Calculating the square root of both sides, we find:
t = √(3.50 / 4.90)
Calculating the value, we get:
t ≈ 0.845 seconds
Therefore, it takes approximately 0.845 seconds for the student to overcome Mr. Horn.
(b) The distance the student travels before catching up with Mr. Horn:
To calculate this, we can substitute the value of 't' we just found into either equation.
Using the distance equation for the student:
Distance covered by the student = (1/2) * 4.90 * (0.845s)^2
Calculating the value, we find:
Distance covered by the student ≈ 1.5892 meters
Therefore, the student travels approximately 1.5892 meters before catching up with Mr. Horn.
(c) The speed of both vehicles at the instant the student overtakes Mr. Horn:
To find the speed of both vehicles, we can calculate their velocities using the equations of motion.
For the student:
Velocity of the student = initial velocity of the student + acceleration of the student * t
Since the student starts from rest, the initial velocity is 0. Therefore,
Velocity of the student = acceleration of the student * t
Substituting the values:
Velocity of the student = 4.90 * 0.845
Calculating the value, we get:
Velocity of the student ≈ 4.1445 m/s
For Mr. Horn:
Velocity of Mr. Horn = initial velocity of Mr. Horn + acceleration of Mr. Horn * (1s)
Since Mr. Horn starts from rest, his initial velocity is 0. Therefore,
Velocity of Mr. Horn = acceleration of Mr. Horn * (1s)
Substituting the values:
Velocity of Mr. Horn = 3.50 * 1
Velocity of Mr. Horn = 3.50 m/s
Therefore, at the instant the student overtakes Mr. Horn, the speed of the student is approximately 4.1445 m/s, and the speed of Mr. Horn is 3.50 m/s.