Answer:
Explanation:
a )
initial velocity u = 45 m/s
acceleration a = - 5 m/s²
final velocity v = 0
v = u - at
0 = 45 - 5 t
t = 9 s
b )
s = ut - 1/2 at²
= 45 x 9 - .5 x5x 9²
405 - 202.5
202.5 m
2 )
a )
s = ut + 1/2 a t²
u = 0
s = 1/2 at²
= .5 x 9.54 x 6.5²
= 201.5 m
b )
v = u + at
= 0 + 9.54 x 6.5
= 62.01 m / s
3
a )
acceleration = (v - u) / t
= (34 - 42) / 2.4
= - 3.33 m /s²
b )
v² = u² - 2 a s
34² = 42² - 2 x 3.33² s
s = 27.41 m
c )
Average velocity
Total displacement / time
= 27.41 / 2.4
= 11.42 m /s
4 )
a )
v = u + at
v = 0 + 3 x 4
= 12 m /s
b )
s = ut + 1/2 a t²
= o + .5 x 3 x 4²
= 24 m
Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.
dr/dt = 4 ft/sec
after 12 seconds, the radius becomes = 
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt
dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
