D. March because it is just below the 1 million marker on the graph and it is the only one that low.
Answer:
a) It takes her 1.43 s to reach a speed of 2.00 m/s.
b) Her deceleration is - 2.50 m/s²
Explanation:
The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:
v = v0 + a · t
Where:
v = velocty.
v0 = initial velocity.
a = acceleration.
t = time.
a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:
v = v0 + a · t
2.00 m/s = 0 m/s + 1.40 m/s² · t
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
It takes her 1.43 s to reach a speed of 2.00 m/s
b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:
v = v0 + a · t
0 = 2.00 m/s + a · 0.800 s
-2.00 m/s / 0.800 s = a
a = -2.50 m/s²
Her deceleration is - 2.50 m/s²
Work= Force x Distance
Answer: 7500 Joules
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + ²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
(1,500 meters) x (1 sec/330 meters) =
(1,500 / 330) (meters-sec/meters) =
4.55 seconds