The working equation would be Vf (final velocity) = Vi
(initial velocity) + a (acceleration) t (time). The given data are the initial
velocity (5.0 m/s), acceleration (-2.5 m/s^2, negative since it is said to
decelerate) and the final velocity (0 m/s, since it will put to a stop). The
time would be 2 seconds.
Answer:
Explanation:
Given:
- area of piston on the smaller side of hydraulic lift,
- area of piston on the larger side of hydraulic lift,
- Weight of the engine on the larger side,
Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.
is the required effort force.
Answer:
(a) the average acceleration of the bus while braking is 8.333 m/s
(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a. [¹/₂ (8.333 m/s) = 4.16 s]
Explanation:
Given;
initial velocity of the bus, v = 25 m/s
time of the motion, t = 3 s
(a) the average acceleration of the bus while braking
a = dv/dt
where;
a is the bus acceleration
dv is change in velocity
dt is change in time
a = 25 / 3
a = 8.333 m/s
(b) If the bus took twice as long to stop, the duration = 2 x 3s
a = 25 / (2 x 3s)
a = ¹/₂ x (25 / 3)
a = ¹/₂ (8.333 m/s) = 4.16 s
Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.
Answer:
Explanation:
Acceleration acts for 9 s and deceleration acts for 12 - 9 = 3 s.
Total distance covered = 990 m
initial velocity u = o
Distance covered while accelerating
s₁ = 1/2 a 9² where a is the acceleration
= 40.5 a
final velocity after 9 s
v = at = 9a
while decelerating
v² = u² - 5 x s₂
0 = (9a)² - 5 s₂
s₂ = 16.2 a²
Distance covered while decelerating = 16.2 a²
s₁ + s₂ = 990
40.5 a + 16.2 a² = 990
16.2 a² + 40.5 a - 990 = 0
a = 6.5
