Question

Acceleration?

Answer 1

initial velocity (vi) is 28 m/s
distance of object = 60 m
reaction time: 0.50 s
vf= 0m/s

Now from what you did:
d=vi(t)+1/2at^2

the value of t in this formula should not be 0.50 s..
if you plugged in all the values and interpret it, it means that the car travels with a distance of 60 m in just 0.50s, which I think is impossible.

You have to find the reaction distance first and then the breaking distance.
Find the reaction distance (prior to breaking):
\(\sf d_{react}= (v_i) (t)\)

Then the breaking distance will be the difference between the distance of the object and the reaction distance that you just calculated.

Using this breaking distance, you can now calculate for the acceleration using this kinematics equation:

\(\sf d_{braking} = \frac{(v_f^2 - v_i^2)}{ (2a)}\)

Answer 2

Okay so here it goes
d react = 28 * .5 = 14
d breaking = 60-14= 46
46 = (0^2- 28^2)/(2a)
46=-784/2a
a= -8.521

Answer 3

that's what I got :)
does it make sense though?

note: I also tend to make mistakes sometimes

Answer 4

Seems like i was clueless from the beginning - but your solution makes sense! thansk a bunch

Answer 5

would u mind checking 2 of my multiple choice problems? (no solving involved)

Answer 6

alright, i can try ^_^

Answer 7

Thanks A LOT! Will attach the files:)

Answer 8

kk

Answer 9

for the first one, is the first graph a vt graph? and the 2nd graph is an at graph?
sorry can't see them properly

Answer 10

yes! (image was blurry for me too)

Answer 11

For the 2nd one, magnitude can't be negative. Negative in vectors refer to their direction. The question only mentioned about the magnitude of the vectors and not their directions.

3rd one, I'll agree with your answer..assuming the ball is under free fall.

Answer 12

so would the second one be choice three

Answer 13

i think so