Amotorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. calculate (a) the acceleration of the motorcycle and (b) its velocity at the beginning and end of the 2-km trip.

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time  t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²    

put value

1000 = v(80) + 1/2×a(80)²           1

and

2000 = v(200) + 1/2×a(200)²      2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

so by kinematic final velocity will be

V² = v² + 2ad

V² = (14.167)² + 2×(-0.042)×(2000)

V²  = 32.70

V = 5.7183 m/s

so

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

false

explanation:

boats, things on boats luggage are way more heavy then water.

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles