1. A train running at 30 m/s is slowed uniformly to a stop in 44s . find the acceleration and stopping distance.

2. a car velocity increases uniformly from 6m/s to 20m/s while covering 70m find the acceleration and the time taken

1. acceleration = -(30 m/s)/44s

= -0.682 m/s^2
stopping distance
= average velocity * (stopping time)
= (15 m/s)*44 s = ___

2. Time taken
= (distance/(average velocity)
= 70 m/13 m/s = 5.38 s

Acceleration = (14 m/s)/5.38 s
= 2.6 m/s^2

1. The train is running at a certain velocity and then being slowed until it stops. So we can conclude here that :

Vi = 30 m/s ; Vf = 0 (because the train stops) ; t=44 s

a) Acceleration (a)?
We can use the formula below:
Vf = Vo + a.t
0 = 30 + a.44
0 = 30 + 44a
-30 = 44a
a = -30/44 = -0,68 (Why negative? because the train slows down, so de-acceleration)

b) Stopping distance (d)?
We can use the formula below :
d = Vi. t + 1/2.a.t^2
d = 30.44 + 1/2 (-0.68) (44)^2
d = 1320 + (-0.68) . 968
d = 1320- 658,24
d = 661,76 m

2. A car's velocity INCREASES, which means it accelerates. So :
Vi =6 m/s ; Vf = 20 m/s ; d=70m

Acceleration (a) ?

We can use the formula below :
Vf^2 = Vi^2 + 2.a.d
(20)^2 = (6)^2 + 2.a.70
400 = 36 + 140a
364 = 140 a
a = 2,6 m/s^2

Note :
Vi : Initial Velocity
Vf : Final Velocity
d : distance taken
a : acceleration

English is not my first language and I apologize if I make any mistakes but I hope it helps!

1. Well, well, well, looks like this train really needed a break! Let's help it out, shall we? So, the initial velocity of the train is 30 m/s, and it comes to a stop in 44 seconds. To find out the acceleration, we can use the good ol' formula: acceleration equals change in velocity divided by time. Since the final velocity is 0 m/s, and the initial velocity is 30 m/s, the change in velocity is -30 m/s. Now we divide that by the time, which is 44 seconds, and voila, we get an acceleration of approximately -0.68 m/s^2. As for the stopping distance, we'll have to use another formula: stopping distance equals initial velocity squared divided by twice the acceleration. Plugging in the values, we find the stopping distance to be approximately 312.35 meters. That's quite a distance to bring the train to a halt! It must have been a very determined train driver!

2. Ah, the need for speed! This car is revving up its engine and going places! So, the initial velocity is 6 m/s, the final velocity is 20 m/s, and the distance traveled is 70 meters. To find the acceleration, we can use the formula: acceleration equals change in velocity divided by time. Since the change in velocity is 20 m/s - 6 m/s = 14 m/s, and the time is unknown for now, we'll have to use another formula: distance equals initial velocity multiplied by time plus half the acceleration times the time squared. Plugging in the values, we get 70 meters = 6 m/s * t + 0.5 * a * t^2. And since the initial velocity is 6 m/s, we know that the first term is just 6t. Now, if we simplify the equation and put it in the form of a quadratic equation, we get 0.5 * a * t^2 + 6t - 70 = 0. Now we can solve this equation to find the value of t. Once we have the value of t, we can substitute it back into the acceleration formula to find the acceleration. I apologize for the lack of an exact answer, but the calculations involved are a bit too complex for this joker!

To solve these problems, we will use the equations of motion for uniformly accelerated motion.

1. To find the acceleration and stopping distance of the train, we can use the equation:

v = u + at

Where:
v = final velocity (0 m/s, because the train comes to a stop)
u = initial velocity (30 m/s)
a = acceleration (unknown)
t = time taken to stop (44 seconds)

Rearranging the equation to solve for acceleration (a), we have:

0 = 30 + a * 44

Solving for 'a', we get:

a = (0 - 30) / 44
a ≈ -0.68 m/s²

This negative value indicates deceleration (slowing down) of the train.

To find the stopping distance, we can use the equation:

s = ut + 1/2 * a * t²

Plugging in the values we know:

s = 30 * 44 + 1/2 * (-0.68) * (44)²
s ≈ 660 meters

Therefore, the acceleration is approximately -0.68 m/s², and the stopping distance is approximately 660 meters.

2. To find the acceleration and the time taken for the car, we can use the following equation:

v² = u² + 2as

Where:
v = final velocity (20 m/s)
u = initial velocity (6 m/s)
a = acceleration (unknown)
s = distance covered (70 meters)

Rearranging the equation to solve for acceleration (a), we have:

a = (v² - u²) / (2s)
a = (20² - 6²) / (2 * 70)
a ≈ 1.72 m/s²

Hence, the acceleration is approximately 1.72 m/s².

To find the time taken (t), we can use the equation:

v = u + at

Rearranging the equation to solve for time (t), we have:

t = (v - u) / a
t = (20 - 6) / 1.72
t ≈ 8.14 seconds

Therefore, the acceleration is approximately 1.72 m/s², and the time taken is approximately 8.14 seconds.

a body falls freely from rest