3 a )Given:
u( initial velocity):60Km/hr=16.67m/sec
v(final velocity):120Km/hr=33.33m/sec
a(acceleration):20 m/s^2
Consider s as the distance traveled by the car. We can calculate s from the below formula.
v^2 - u^2= 2as
Where v is the final velocity measured in m/s
u is the initial velocity measured in m/s
a is the acceleration measured in m/s^2
s is the distance traveled by the car.
Substituting the given values in the above formula we get
33.33^2- 16.67^2= 2 x 20 x s
832.99= 40 s
s = 250.12 m
3b) Consider t as the time taken for the car to travel the above distance. We can calculate t from the below formula.
s = ut +1/2(at^2)
250.12= 16.67 X t + 1/2(20 x t^2)
500.233 = 33.33t + 20t^2
Solving the above quadratic equation we get t= 1026 secs.
4) Given
v( final velocity) = 0.
time taken to cover the distance= 25 secs
Distance traveled(s)=40Km= 40000m
Now consider the below equation
v = u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
Substituting the given values in the above equation we get
0= u+ax25
u= -25a
Now we already know that
s = ut + 1/2(at^2)
Where s is the distance traveled
u is the initial velocity
a is the acceleration
t is the time
Substituting the given values in the above formula we get
40000 = u25 +1/2(ax25x25)
Now as solved above -25a =u. Substituting this in the above formula we get
40000= 25u +1/2(-25u)
40000= 12.5u
Thus u = 40000/12.5
u = 3200m/s
As per the above derived equation
We know in this case
-25 a = u
a= -128 m/s^2
