Question

S. An armadillo scurries along in a straight line (the a-axis). The equation for the armadillo's position vector is ) 150.0em i+(12 0cm/a)t i -(0.480cm/s? i e. Calculate when the armadillo turns around and starts to scurry back the way it came f. Calculate the time that elapses (starting from t0) until the armadillo returns to its starting position. Answer: 25 s. g. Calculate the velocity of the armadillo when it returns to its starting position. * h. Calculate the time it takes for the armadillo to reach a position of 200.0 cm after it has turned around

Answer 1

Firstly, let's find equations for v and a

x = 150 + 12t - 0.48t^2

v = dx/dt = 12 - 0.96t

a = dv/dt = -0.96

e)

When armadillo is just turning, it stops momentarily. Hence, v = 0

0 = 12 - 0.96 t

t = 12.5 s

f)

If it returns to same position, x at t= 0

x = 150 at t = 0.

Putting x = 150 again

150 = 150 + 12t - 0.48t^2

solving the quadratic eqn we get = 25 sec

g)

v at t = 25 sec

v = 12 - 0.96 x 25 = -12 cm/s ... - ve sign means going backwards

h)

put x = 200 and solve the quadratic equation. But only consider time greater than 12.5 sec as answer because it is asked the time after turning around

200 = 150 + 12t - 0.48t^2

t = 19.7 sec