Question

Solve it ...........?​

Answer 1

Answer:

HERE IS YOUR SOLUTION WITH THE ATTACHMENT.....

Answer 2

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf a=4\sqrt{2}m/{s}^{2}}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Radius (r) = 1m

• Speed v = $\sf{(2{t}^{2})m/s}$

• Time (t) = 1s

$\large\underline\pink{\sf To\:Find: }$

• Magnitude of acceleration or Net acceleration = ?

━━━━━━━━━━━━━━━━━━━━━━━━━

We know that :-

$\large{♡}$$\large{\boxed{\sf Tangential\: Acceleration (a_t)={\frac{dv}{dt}}}}$

$\large\implies{\sf 4t }$

at t = 1s

$\large{\sf a_t=4m/{s}^{2}}$

________________________________________

$\large{♡}$$\large{\boxed{\sf Centripetal\: Acceleration (a_C)={\frac{{v}^{2}}{r}}}}$

$\large\implies{\sf {\frac{4{t}^{4}}{1}}}$

at t = 1s

$\large{\sf a_C=4\sqrt{2}m/{s}^{2}}$

_______________________________________

Net Acceleration (a) :

$\large{\boxed{\sf a =\sqrt{a_t^{2}+a_C^{2}}}}$

$\large\implies{\sf \sqrt{{4}^{2}+{4}^{2}}}$

$\huge\red{♡}$$\large\red{\boxed{\sf a=4\sqrt{2}m/{s}^{2}}}$