Question

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

Answer 1

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

Answer 2

a body falls freely from rest

Answer 3

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!