Question

A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.

Answer 1

Answer:

a)$a=5 i+2t j - 6\ t^2k$

b)$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k$

$a=5 i+2t j - 6\ t^2k$

Therefore the acceleration function a will be

$a=5 i+2t j - 6\ t^2k$

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

$a=\sqrt{5^2+4^2+24^2}\ m/s^2$

a= 24.83 m/s²

The direction of the acceleration a is given as

$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

a)$a=5 i+2t j - 6\ t^2k$

b)$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$