A car starts from rest and accelerates uniformaly at the rate of 0.98m/s in 1hr.what is its final velocity?

A car starts from rest and reaches a velocity of 40m/s in 1minutes..(a)what is its acceleration?(b) If its acceleration remains constant,what will its velocity be in 15 minutes?
The brakes of a certain car can produce an acceleration of 6m/s2.(a)how long does it take the car to come to a stop from a velocity of 30m/s?(b) how far does the car travel during the time the brakes are applied?
A train is uniformly decelerated at 2.5 m/s2 during a 2s period.how far does it travel before coming to a full stop?
A cyclist travelling uniformly at 30m/s sees a post 300m away and suddenly applies his brakes.If the bike stops in 4.8s,determine whether the cyclist hits the post.
A boy increases speed from 1.2m/s to 2.4m/s in 6s,how far does the boy travel?

1 hr = 3600s.

Vf = Vo + at,
Vf = 0 + 0.98*3600 = 3528m/s.

a. a = (Vf - Vo) / t,
a = (40 - 0) / 60s = 0.67m/s^2.

b. t = 15 min. * 60s/min = 900s.
Vf = Vo + at,
Vf = = 0 + 0.667 * 900 = 600m/s.

a. t = (Vf - Vo) / a,
t = (0 - 30) / -6 = 5s.

b. d = (Vf^2 - Vo^2) / 2a,
d = (0 - (30)^2) / -12 = 75m.

Vf = Vo + at,
Vo = Vf - at,
Vo = 0 - (-2.5)*2 = 5m/s.

d = (Vf^2 - Vo^2) / 2a,
d = (0 - (5)^2) / -5 = 5m.

Correction: Disregard the 1st 3 lines,

and start at "a".

To answer these questions, we can use the equations of motion related to uniform acceleration:

1. Final velocity (v) can be calculated using the formula:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

2. Acceleration (a) can be calculated using the formula:
a = (v - u) / t

Now let's proceed to solve each question:

1. The car starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is given as 0.98 m/s², and the time (t) is 1 hour = 3600 seconds.
Using the formula v = u + at:
v = 0 + (0.98 * 3600)
v = 3528 m/s
Therefore, the final velocity of the car is 3528 m/s.

2a. The car starts from rest, so the initial velocity (u) is 0 m/s. The final velocity (v) is given as 40 m/s, and the time (t) is 1 minute = 60 seconds.
Using the formula a = (v - u) / t:
a = (40 - 0) / 60
a = 0.67 m/s²
Therefore, the acceleration of the car is 0.67 m/s².

2b. The acceleration remains the same, so we can use the same acceleration value as in part 2a. The time (t) is given as 15 minutes = 900 seconds.
Using the formula v = u + at:
v = 0 + (0.67 * 900)
v = 603 m/s
Therefore, the velocity of the car after 15 minutes will be 603 m/s.

3a. The car is coming to a stop, so the final velocity (v) is 0 m/s. The initial velocity (u) is given as 30 m/s, and the acceleration (a) is given as 6 m/s².
Using the formula v = u + at:
0 = 30 + (6 * t)
Solving for t:
6t = -30
t = -30 / 6
t = -5 seconds
Since time cannot be negative, this means the car does not come to a stop from a velocity of 30 m/s within the given conditions.

3b. To calculate the distance traveled during the time the brakes are applied, we can use the formula:
v² = u² + 2as
where s is the distance.

Rearranging the equation, we have:
s = (v² - u²) / (2a)
Plugging in the given values:
s = (0 - 30²) / (2 * -6)
s = 900 / -12
s = -75 meters
Since distance cannot be negative, the car travels 75 meters during the time the brakes are applied.

4. The train is decelerated, so the acceleration (a) is given as -2.5 m/s². The time (t) is given as 2 seconds.
Using the formula v = u + at:
0 = u + (-2.5 * 2)
-2u = -5
u = 2.5 m/s (initial velocity)

To find the distance traveled, we can use the formula:
s = ut + (1/2)at²
Plugging in the values:
s = 2.5 * 2 + (1/2) * (-2.5) * (2)²
s = 5 - 5
s = 0 meters
Therefore, the train does not travel any distance before coming to a full stop.

5. The cyclist is traveling at an initial velocity (u) of 30 m/s. The final velocity (v) is 0 m/s, and the time (t) is given as 4.8 seconds.
Using the formula v = u + at:
0 = 30 + a * 4.8
-30 = 4.8a
a = -6.25 m/s²
The negative sign indicates deceleration.

To find the distance traveled, we can use the formula:
s = ut + (1/2)at²
Plugging in the values:
s = 30 * 4.8 + (1/2) * (-6.25) * (4.8)²
s = 144 - 57.6
s = 86.4 meters
Therefore, the cyclist travels 86.4 meters before coming to a stop.

I hope this explanation helps! If you have any further questions, feel free to ask.