The car is travelling along the road with a speed of v (300 / s) m/s, where s is in
meters. Determine the magnitude of its acceleration when t = 3 s if t = 0 at s = 0. ( ans: a
= 1.28 m/s2).
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v = ds/dt
ds/dt = 300/s
300dt = s ds then integrate both side (0<t<3) & (0<s<S)
To determine the magnitude of acceleration, we can use the formula:
acceleration = change in velocity / change in time
Given that the car's speed is v = 300 / s m/s, we need to find the change in velocity as t changes from 0 to 3 seconds.
At t = 0, s = 0, which means the initial velocity is:
v_initial = 300 / 0 m/s (undefined)
At t = 3, the velocity is:
v_final = 300 / 3 m/s = 100 m/s
Now, we can calculate the change in velocity:
change in velocity = v_final - v_initial = 100 - undefined
Since we have an undefined initial velocity, we need to find the limit of the velocity as s approaches 0.
lim(s→0) (300 / s) = ∞
Therefore, we can conclude that as t approaches 0, the speed of the car approaches infinity.
Due to this, it is not possible to determine the magnitude of acceleration when t = 3 s.
To determine the magnitude of the car's acceleration at t = 3s, we need to find the derivative of the velocity function with respect to time.
Given that the speed, v, is defined as v = (300 / s) m/s, where s is in meters, we need to convert the speed function to a function of time.
Let's recall that speed is the magnitude of velocity, which is the derivative of position with respect to time.
To convert the speed function to a function of time, we need to find the relationship between s and t. Given that t = 0 when s = 0, we can set up a relationship between the two variables.
Let's assume that s = f(t). Therefore, s at t = 3s is f(3).
Now, let's find the relationship between s and t. We know that the car's speed is given by v = (300 / s). Since v = ds/dt (derivative of s with respect to t), we can write the following equation:
ds/dt = 300 / s
To solve this differential equation, we'll multiply both sides by s and dt:
s * ds = 300 * dt
Now, we can integrate both sides of the equation:
∫ s * ds = ∫ 300 * dt
This leads to:
(s^2) / 2 = 300t + C1 (where C1 is the constant of integration)
Since t = 0 when s = 0, we can substitute these values:
(0^2) / 2 = 300 * 0 + C1
C1 = 0
Therefore, the equation becomes:
(s^2) / 2 = 300t
Now, let's solve for s:
s^2 = 600t
Taking the square root of both sides:
s = √(600t)
Now, we have the relationship between s and t. To determine the magnitude of acceleration at t = 3s, we need to find the derivative of velocity with respect to time.
Let's start by finding the derivative of v with respect to t:
v = 300 / s
Taking the derivative with respect to t:
dv/dt = (d(300 / s) / dt)
Using the chain rule, we get:
dv/dt = (-300 / s^2) * (ds / dt)
We already found that ds / dt = √600, so we can substitute it into the equation:
dv/dt = (-300 / s^2) * √600
Now, to find the magnitude of acceleration at t = 3s, we substitute s = √(600t) and t = 3 into the equation:
a = (-300 / (√(600t))^2) * √600
a = (-300 / 600t) * √600
a = -0.5 / t * √600
Substituting t = 3 into the equation:
a = -0.5 / 3 * √600
a ≈ -0.28 * √600
Calculating this value gives:
a ≈ -0.28 * 24.49
a ≈ -6.8772 m/s^2
Since we are asked for the magnitude of acceleration, we take the absolute value:
|a| ≈ 6.8772 m/s^2
Therefore, the magnitude of the car's acceleration at t = 3s is approximately 6.8772 m/s^2.