Question

Find the components of 6 x -7 please answer me fast who answered will mark as brainleist​

Answer 1

Answer:

Step-by-step explanation:

We can find the tangential accelration by using Chain Rule to rewrite the velocity vector as follows:

v=drdt=drdsdsdt=Tdsdt(2.6.1)

Now, since acceleration is simply the derivative of velocity, we find that:

a=dvdt=ddt(Tdsdt)=d2sdt2T+dsdtdTdt=d2sdt2T+dsdt(dTdsdsdt)=d2sdt2T+dsdt(κNdsdt)=d2sdt2T+κ(dsdt)2N(2.6.2)(2.6.3)(2.6.4)(2.6.5)(2.6.6)(2.6.7)

Note

dTds=κN(2.6.8)

This, in turn, gives us the definition for acceleration by components.

Definition: acceleration vector

If the acceleration vector is written as

a=aTT+aNN,(2.6.9)

then

aT=d2sdt2=ddt|v|  and  aN=κ(dsdt)2=κ|v|2(2.6.10)

To calculate the normal component of the accleration, use the following formula:

aN=|a|2−a2T−−−−−−−√(2.6.11)

We can relate this back to a common physics principal-uniform circular motion. In uniform circulation motion, when the speed is not changing, there is no tangential acceleration, only normal accleration pointing towards the center of circle. Why do you think this is? Hint: look in the introduction section for the difference between the two components of acceleration.

Example 2.6.1

Without finding T and N, write the accelration of the motion

r(t)=(cost+tsint)i^+(sint−tcost)j^(2.6.12)

for t>0.

To solve this problem, we must first find the particle's velocity.

v=drdt=(−sint+sint+tcost)i^+(cost−cost+tsint)j^=(tcost)i^+(tsint)j^(2.6.13)(2.6.14)(2.6.15)

Next find the speed.

|v|=t2cos2t+t2sin2t−−−−−−−−−−−−−−√=t2−−√=|t|(2.6.16)

When t>0, |t| simply becomes t.

We know that aT=ddt|v|, which we can use to find that ddt(t)=1.

Now that we know aT, we can use it to find aN using Equation 2.6.11.

a=(cost−tsint)i^+(sint+tcost)j^(2.6.17)

|a|2=t2+1(2.6.18)

aN=(t2+1)−(1)−−−−−−−−−−−√=t(2.6.19)

By substituting this back into the original definition, we find that

|a|=(1)T+(t)N=T+tN(2.6.20)

Example 2.6.2

Write ain the form a=aTT+aNN without finding T or N.

r(t)=(t+1)i^+2tj^+t2k^(2.6.21)

v=ti^+2j^+2tk^(2.6.22)

|v|=5+4t2−−−−−−√(2.6.23)

aT=12(5+4t2)−12(8t)(2.6.24)

=4t(5+4t2)−12(2.6.25)

aT(1)=49–√=43(2.6.26)

a=2k^(2.6.27)

a(1)=2k^(2.6.28)

a(t)=2k^(2.6.29)

Now we use Equation 2.6.11:  

aN=|a|2−a2T−−−−−−−√=22−(43)2−−−−−−−−−√=209−−−√=25–√3(2.6.30)(2.6.31)(2.6.32)

a(1)=43T+25–√3N(2.6.33)