Question
Sally is driving along a straight highway in her classic 1965 Mustang. At time t = 0, Sally is moving at 10 m/s in the positive x-direction, and she passes a signpost at x = 50m. Her x-acceleration is a function of time given by ax(t) = 2.0 m/s2 – 0.10 m/s3 • t
a) Find her x-velocity and position as functions of time?
b) When is her x-velocity greatest?
c) What is the maximum x-velocity?
d) Where is the car when it reaches the maximum x-velocity?
Solution
Data Breakdown:
Sally’s motion has a time-dependent acceleration given by:
ax(t) = 2.0 m/s2 – 0.10 m/s3 • t
Initial conditions:
vx(0) = 10 m/s
x(0) = 50 m
We need to determine:
a) vx(t) and x(t) as functions of time.
b) The time when vx(t) is greatest.
c) The maximum vx.
d) The position x of the car when vx is maximum.
a) Finding vx(t) and X(t):
Step 1: Find vx(t): ax(t) = dvx/dt,
and integrating the ax(t) gives: vx(t) = ∫ax(t)dt = ∫ (2.0 – 0.10t)dt,
Hence, vx(t) = 2.0t – 0.05t2 + C
From the initial condition, vx(0) = 10:
10 = 2.0(0) – 0.05(0)2 + C
Thus, C = 10
This implies that: vx(t) = 2.0t – 0.05t2 + 10
Step 2: Find x(t):
Velocity is the derivative of position
vx(t) = dx/dt
Integrating vx(t) gives: x(t) = ∫vx(t)dt = ∫(2.0t – 0.05t2 + 10)dt
Therefore; x(t) = (2.0t2/2) – (0.05t3/3) + 10t + C
x(t) = t2 + 0.0167t3 + 10t + C
When x(0) = 50
50 = (0)2 + 0.0167(0)3 + 10(0) + C
Hence; C = 50
Thus; x(t) = t2 + 0.0167t3 + 10t + 50
b) When Vx(t) is greatest?
To find the time when vx(t) is greatest, compute when the derivative of vx(t) equals zero:
dvx/dt = ax(t) = 2.0 – 0.10t
Set ax(t) = 0:
2.0 – 0.10t = 0 ⟹ t = 20 seconds
Thus, vx(t) is greatest at t = 20s.
c) What is the maximum vx?
Substitute t = 20s into vx(t):
vx(20) = (2.0)20 – 0.05(20)2 + 10
vx(20) = (2.0)20 – 0.05(400) + 10
vx(20) = 40 – 20 + 10 = 30 m/s.
Thus, the maximum vx is 30 m/s.
(d) Where is the car when vx(t) is maximum?
Substitute t = 20 s into x(t):
x(20) = 202 + 0.0167(20)3 + 10(20) + 50
x(20) = 400 + 0.0167(8000) + 200 + 50
x(20) = 400 + 133.6 + 200 + 50 = 516.4 m
Thus, the car is at x = 516.4 m when vx(t) is maximum.
Final Answers:
(a) The velocity and position as functions of time are:
vx(t) = 2.0t – 0.05t2 + 10, x(t) = t2 – 0.0167t3 + 10t + 50
(b) The x-velocity is greatest at t=20s.
(c) The maximum x-velocity is 30 m/s.
(d) The car is at x = 516.4 m when vx is maximum.