Answer:
a. 
b. 
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c. 
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,
- speed of lagging car,
- distance between the cars,
- deceleration of the leading car after braking,
a.
Time taken by the car to stop:
where:
, final velocity after braking
time taken
b.
using the eq. of motion for the given condition:
where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:
is the time taken to stop after braking
The answer will be
(1) correct
(2) correct
(3) the force of the soccer ball on the net
(4) Will not change
Hope this help
Answer:
Where is the graph??
If a car travels from zero to 4 m/s ins 8 sec
a = 4 / 8 = .5 m/s^2
V (2) = 2 * .5 = 1 m/s after 2 sec
S = V t + 1/2 a t^2
S = 1 * 3.9 + 1/2 * 1/2 *3.9^2 = 3.9 + 3.80 = 7.70 m from 2 to 5.9 sec
Check:
Total distance traveled = 1/2 a t^2 = 5.9^2 / 4 = 8.70 m
Distance traveled in 2 sec = 1/2 * 1/2 * 4 = 1 m
Total distance from 2 to 5.9 = 8.7 - 1 = 7.7 m agreeing with thw above
Answer:
(D)
Explanation:
A(2,3,4,6,8)
B(6,9,12,15,18)
(AuB)=(2,3,4,6,8,9,12,15,18)
Would that be 10?
.1 second per revolution, so 10 rps
