Question

Please help!!! An object is moving initially with a velocity of 5.1 m/s . After 3.1 s the object's velocity is -2.7 m/s . What is the object's average acceleration during this time?

Answer 1

Answer:

Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,

d

d

t

v

(

t

)

=

a

(

t

)

,

ddtv(t)=a(t),

we can take the indefinite integral of both sides, finding

∫

d

d

t

v

(

t

)

d

t

=

∫

a

(

t

)

d

t

+

C

1

,

∫ddtv(t)dt=∫a(t)dt+C1,

where C1 is a constant of integration. Since

∫

d

d

t

v

(

t

)

d

t

=

v

(

t

)

∫ddtv(t)dt=v(t), the velocity is given by

v

(

t

)

=

∫

a

(

t

)

d

t

+

C

1

.

v(t)=∫a(t)dt+C1.

Similarly, the time derivative of the position function is the velocity function,

d

d

t

x

(

t

)

=

v

(

t

)

.

ddtx(t)=v(t).

Thus, we can use the same mathematical manipulations we just used and find

x

(

t

)

=

∫

v

(

t

)

d

t

+

C

2

,

x(t)=∫v(t)dt+C2,

where C2 is a second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find

v

(

t

)

=

∫

a

d

t

+

C

1

=

a

t

+

C

1

.

v(t)=∫adt+C1=at+C1.

If the initial velocity is v(0) = v0, then

v

0

=

0

+

C

1

.

v0=0+C1.

Then, C1 = v0 and

v

(

t

)

=

v

0

+

a

t

,

v(t)=v0+at,

which is (Equation). Substituting this expression into (Figure) gives

x

(

t

)

=

∫

(

v

0

+

a

t

)

d

t

+

C

2

.

x(t)=∫(v0+at)dt+C2.

Doing the integration, we find

x

(

t

)

=

v

0

t

+

1

2

a

t

2

+

C

2

.

x(t)=v0t+12at2+C2.

If x(0) = x0, we have

x

0

=

0

+

0

+

C

2

;

x0=0+0+C2;

so, C2 = x0. Substituting back into the equation for x(t), we finally have

x

(

t

)

=

x

0

+

v

0

t

+

1

2

a

t

2

,

x(t)=x0+v0t+12at2,