Question

Please i need answer ASAP... Please guys

Answer 1

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