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A rabbit is moving in the positive x-direction at 1.50 m/s when it spots a predator and accelerates to a velocity of 12.1 m/s along the positive y-axis, all in 2.20 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.)

User Smaran
by
6.4k points

1 Answer

4 votes

Answer:

Explanation:

First we have to consider the initial and final velocities to calculate the average acceleration:

v_i= - \Big[ 1.10\frac{m}{s}\Big]\widehat{i} \\ v_f= \Big[ 11.5\frac{m}{s}\Big]\widehat{j} \\ \implies \Delta v=v_f-v_i \\ \implies \Delta v= [ 1.10\frac{m}{s}]\widehat{i} +[ 11.5\frac{m}{s}]\widehat{j} \text{ in an average time } \Delta t=1.40\,s.v

i

=−[1.10

s

m

]

i

v

f

=[11.5

s

m

]

j

⟹Δv=v

f

−v

i

⟹Δv=[1.10

s

m

]

i

+[11.5

s

m

]

j

in an average time Δt=1.40s.

With the information we can calculate the average acceleration in terms of the x- and y-components:

a=\Delta v/\Delta t= \dfrac{[ 1.10\frac{m}{s}]\widehat{i} +[ 11.5\frac{m}{s}]\widehat{j}}{1.40\,s} \\ \text{ } \\ \implies a= \big[ 0.786\frac{m}{s^2}\big]\widehat{i} + \big[ 8.214\frac{m}{s^2} \big]\widehat{j}a=Δv/Δt=

1.40s

[1.10

s

m

]

i

+[11.5

s

m

]

j

⟹a=[0.786

s

2

m

]

i

+[8.214

s

2

m

]

j

User Dima Tisnek
by
6.0k points