A motor cycle is being driven at a speed of 20 m/s when brakes are applied to bring it to rest in five seconds the deccerlation produced in case is?
Answers
In the above Question, the following information is given -
A motorcycle is being driven at a speed of 20 m/s.
When brakes are applied, it comes to rest in five seconds.
To find -
The de-acceleration produces in this case.
Solution -
Here, the initial speed of the motorcycle is 20 m / s.
It finally comes to rest.
So, the final speed of the motorcycle becomes 0 m / s.
Now,
The time is taken for the motorcycle to come to rest in 5 seconds.
Now, from the first equation of motion,
We know that -
V = u + at
Where -
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time taken.
Substituting the given values into this equation -
0 = 20 + 5a
=> 5a = -20
=> a = -4 m / s²
Here, as the value of acceleration is negative, retardation is taking place.
____________
Given that, motorcycle is moving with a speed of 20 m/s (means the initial velocity of the motorcycle is 20 m/s) when brakes are applied it comes to rest in five seconds.
We have to find the deccerlation produced.
From above data we have; initial velocity i.e. u is 20 m/s, final velocity i.e. v is 0 (as brakes are applied) and time i.e. t is 5 sec.
First equation of motion:
v = u + at
Second equation of motion:
s = ut + 1/2 at²
Third equation of motion:
v² - u² = 2as
As per given condition we have value of v, u and t. So,
Using the First Equation Of Motion,
v = u + at
Substitute the known values in the above formula,
0 = 20 + a(5)
-20 = 5a
Divide by 5 on both sides,
-20/5 = 5a/5
-4 = a
(Negative sign shows retardation or deceleration)
Therefore, the acceleration produced is 4 m/s².