Determine the car's acceleration when t=4sec. ANS: 1.06 m/sec^2

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In summary: So in this case, we have x = s^{-\frac{1}{3}} and dx = dsPutting these values in the formula, we get:\int s^{-\frac{1}{3}} ds = \frac{s^{\frac{2}{3}}}{\frac{2}{3}} + cSince we are dealing with indefinite integrals, we need to add a constant c at the end. And since we have already been given the value of s at t=0 as 0, we can assume c=0.Hence, we get \frac{3}{2} s^{\frac{2}{3
  • #1
jjiimmyy101
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Problem: A car starts from rest and moves along a straight line with an acceleration of a=(3s^-1/3)m/sec^2, where s is in metres. Determine the car's acceleration when t=4sec. ANS: 1.06 m/sec^2

Alright...I know nothing about integrals...really, nothing. I was never taught anything about integrals even though I've taken calculus courses before.

Here's what I think I should do.

Take the equation a = d^2s/dt^2 and INTEGRATE it to find the position (s) and then substitute back into original equation. But how do you do this.
 
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  • #2
If u want to taste calculus here it goes then

Now [tex]a=\frac{dv}{dt}=\frac{dv*ds}{ds*dt}=\frac{vdv}{ds}=3s^{-\frac{1}{3}}[/tex]

[tex] vdv=3s^{-\frac{1}{3}} ds[/tex]
integrate u wll get [tex] v^2=9s^{\frac{2}{3}}+c [/tex]

From conditions given c=0
therefore [tex] v^2=9s^{\frac{2}{3}} [/tex]
Now [tex] v=3s^{\frac{1}{3}}[/tex]

v=ds/dt

so we again have

[tex] s^{-\frac{1}{3}} ds = 3dt[/tex]
Again integrating u get
[tex]\int s^{-\frac{1}{3}} ds = \int 3dt[/tex]
u get
[tex]\frac{3}{2} s^{\frac{2}{3}}=3t+c[/tex]

From the given conditions c=0
so we have
[tex]s^{\frac{2}{3}}=2t[/tex]

So at t=4, s=[tex]8^{\frac{3}{2}}[/tex]

and hence acceleration a= 3[tex]8^{-\frac{1}{3}}[/tex] = 1.06
 
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  • #3
Thank-you.

I'll apologize before I even ask.
Sorry.

I know you integrated, but could you show me in more detail how you went from v*dv = 3*s^-1/3*ds
to
v^2 = 9*s^2/3 + c
 
  • #4
It is a basic formula

[tex]\int x^n dx = \frac{x^{n+1}}{n+1}[/tex]
 

1. What is the formula for determining acceleration?

The formula for acceleration is acceleration (a) = change in velocity (v) / change in time (t).

2. How do you determine the car's acceleration when given a specific time?

To determine the car's acceleration when given a specific time, you need to know the change in velocity and change in time at that specific time. Plug these values into the formula a = v/t and you will get the acceleration at that particular time.

3. What unit is used to measure acceleration?

Acceleration is measured in meters per second squared (m/sec^2).

4. How do you interpret a negative acceleration value?

A negative acceleration value indicates that the car is decelerating, or slowing down, at that specific time.

5. Is the acceleration value constant or changing over time?

The acceleration value can be constant or changing over time, depending on the car's motion. If the car is moving at a constant velocity, the acceleration value will be 0. If the car is accelerating or decelerating, the acceleration value will be changing over time.

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