Question

1) Find the x component of the average velocity of the asteroid.
2) Find the magnitude of the average acceleration of the asteroid.

The velocity as a function of time for an asteroid in the asteroid belt is given by where vo and , are constants Use r =0 as the initial time, and =1.8 as the final time. Hint: Doing your work algebraically first as you always should you will find that being given the final time this way will cut down a lot of number crunching in your calculator! The values for the constants that you will use are: 3.5 m/s 882s

Answer 1

$\vec{v(t)}= v_{o}e^{\frac{-t}{t_{o}}}\hat{i} + \frac{v_{o}t}{2t_{o}}\hat{j}$

given

v_o =3.5ms^-1

t_o=882s

Solution

1)

avg velocity along x direction= total distance travelled along x direction/ total time

Now, velocity along x direction

$\vec{v(x)}= v_{o}e^{\frac{-t}{t_{o}}}$

$\Rightarrow \frac{\mathrm{d}x }{\mathrm{d} t}= v_{o}e^{\frac{-t}{t_{o}}}$

$\Rightarrow \mathrm{d}x = v_{o}e^{\frac{-t}{t_{o}}}\mathrm{d} t$

$\Rightarrow \int_{0}^{x}\mathrm{d}x = v_{o}\int_{0}^{1.8t_{o}}e^{\frac{-t}{t_{o}}}\mathrm{d} t$

$\Rightarrow x= \left |-v_{o}t_{o}e^{\frac{-t}{t_{o}}} \right |_{0}^{1.8t_{o}}$

$\Rightarrow x= -3.5*882e^{-1.8}} m$

$\Rightarrow x= -3087e^{-1.8}} m$

$\Rightarrow x= 510.28 m$, which is the total displacement

total time= t_f -t_i

T=1587.6s

Average velocity along x axis is

$\frac{x}{T}= \frac{510.28}{1587.6} ms^{-1}= 0.3214ms^{-1}$

2)

Average acceleration is given by change in velocity/ time taken

Velocity at t=0

$\vec{v}(0)= v_{o}\hat{i}$

$\Rightarrow \vec{v}(0)= 3.5\hat{i}$

Velocity after time t=1.8t_o

$\vec{v}(1.8t_{o})= v_{o}e^{-1.8}\hat{i} + \frac{v_{o}*1.8}{2}\hat{j}$

$\Rightarrow \vec{v}(1.8t_{o})= 3.5e^{-1.8}\hat{i} + 3.15\hat{j}$

$\Rightarrow \vec{v}(1.8t_{o})= 3.5*0.165\hat{i} + 3.15\hat{j}$

$\Rightarrow \vec{v}(1.8t_{o})= 0.58\hat{i} + 3.15\hat{j}$

Now, avg acceleration

$\Rightarrow \vec{a}_{avg}= \frac{0.58\hat{i} + 3.15\hat{j}-3.5\hat{i}}{1.8*882}$

$\Rightarrow \vec{a}_{avg}= \frac{-2.92\hat{i} + 3.15\hat{j}}{1587.6}$

$\Rightarrow \left |\vec{a}_{avg} \right |= \frac{\sqrt{\left (-2.92 \right )^{2} + 3.15^{2}}}{1587.6}$

$\Rightarrow \left |\vec{a}_{avg} \right |= \frac{4.3}{1587.6}ms^{-2}$

$\Rightarrow \left |\vec{a}_{avg} \right |=2.7*10^{-3}ms^{-2}$ which is the required answer

If any doubt feel free to comment