Each of four particles moves along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by:
Particle #1: x(t)=2t^2-4
Particle #2: x(t)= -3t-4
Particle #3: -2t^2-4
Particle #4: 2 *[sq. root (t+1)] -4
Which of these particles are speeding up for 0<t< infinity?
a)#1 only
b)#1,3,4
c)all four
d)none
e)#1,4
To determine which particles are speeding up for 0<t< infinity, we need to find the acceleration of each particle. The acceleration of a particle can be found by taking the second derivative of its position function with respect to time.
Let's calculate the acceleration for each particle:
Particle #1:
x(t)=2t^2-4
To find the acceleration, we take the derivative of the position function twice:
a(t) = d^2x(t)/dt^2
= d^2(2t^2-4)/dt^2
= d/dt(4t)
= 4
Particle #2:
x(t)=-3t-4
Taking the derivative twice:
a(t) = d^2x(t)/dt^2
= d^2(-3t-4)/dt^2
= d/dt(0)
= 0
Particle #3:
x(t)=-2t^2-4
Taking the derivative twice:
a(t) = d^2x(t)/dt^2
= d^2(-2t^2-4)/dt^2
= d/dt(-4)
= 0
Particle #4:
x(t) = 2 * √(t+1) - 4
To find the acceleration, we take the derivative of the position function twice:
a(t) = d^2x(t)/dt^2
= d^2(2√(t+1) - 4)/dt^2
= d/dt(0)
= 0
Now we have the accelerations of each particle:
Particle #1: a(t) = 4
Particle #2: a(t) = 0
Particle #3: a(t) = 0
Particle #4: a(t) = 0
Particles that are speeding up have positive accelerations. From the calculations, we can see that only Particle #1 has a positive acceleration (a(t) = 4). Therefore, the answer is:
a) #1 only
To determine which particles are speeding up, we first need to find the velocity functions for each particle by taking the derivative of the position function with respect to time.
Particle #1: x(t) = 2t^2 - 4
Velocity: v(t) = d/dt (2t^2 - 4) = 4t
Particle #2: x(t) = -3t - 4
Velocity: v(t) = d/dt (-3t - 4) = -3
Particle #3: x(t) = -2t^2 - 4
Velocity: v(t) = d/dt (-2t^2 - 4) = -4t
Particle #4: x(t) = 2√(t + 1) - 4
Velocity: v(t) = d/dt (2√(t + 1) - 4) = 2/(2√(t + 1)) = 1/√(t + 1)
To determine if a particle is speeding up, we look at the sign of its velocity. If the velocity is positive, the particle is moving in the positive direction (speeding up), and if the velocity is negative, the particle is moving in the negative direction (slowing down).
For Particle #1, the velocity is 4t. Since t is positive for 0 < t < infinity, the velocity is always positive, indicating that Particle #1 is speeding up.
For Particle #2, the velocity is -3. The velocity is always negative, indicating that Particle #2 is slowing down.
For Particle #3, the velocity is -4t. Since t is positive for 0 < t < infinity, the velocity is always negative, indicating that Particle #3 is slowing down.
For Particle #4, the velocity is 1/√(t + 1). Since the term √(t + 1) is always positive for 0 < t < infinity, the velocity is always positive, indicating that Particle #4 is speeding up.
Therefore, the particles that are speeding up for 0 < t < infinity are Particle #1 and Particle #4.
The answer is e) #1,4.
If you consider "speeding up" to be an increase in speed, regardless of direction, then none of the choices is correct.
#2 and #3 accelerate, backwards. The others start out by slowing down.
They may be referring to increasing speed as an increase in V in the positive direction.