The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:
d = vt = (22 m/s)(12 s) = 264 m
For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²
Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m
Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.
Answer:
3.0883 x 10^10mg
Explanation:
1 kilogram = 1000 000 milligrams
So, 30 883 x 1000 000 = 30 883 000 000mg
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:(a)9.685 mm
(b)4.184 mm
Explanation:
Given
Wavelength of light 
Width of slit(b)=0.210
(a)Width of central maximum located 1.80m from slit


=9.685 mm
(b)Width of the first order bright fringe



Energy is the capacity to do some type of work