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

[jjiimmyy101]

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.

 

[himanshu121]

If u want to taste calculus here it goes then

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

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

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

v=ds/dt

so we again have

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

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

So at t=4, s=8^{\frac{3}{2}}

and hence acceleration a= 38^{-\frac{1}{3}} = 1.06

 

[jjiimmyy101]

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

 

[himanshu121]

It is a basic formula

\int x^n dx = \frac{x^{n+1}}{n+1}