A bus moving at a speed of 30ms^-1 is brought to rest within 10s by applying brakes. What is the retardation? Also calculate the distance travelled by bus before coming to rest?
Answers
Given :
Initial velocity : 30 m/s
Final velocity : 0 m/s
Time taken : 10 s
To find :
The retardation and the distance travelled.
Solution :
As we are provided with initial velocity, final velocity and time taken we can use 1st equation of motion that is,
v = u + at
where, v denotes final velocity, u denotes initial velocity, a denotes acceleration and t denotes time taken.
By substituting all the given values in the equation,
➙ v = u + at
➙ 0 = 30 + a(10)
➙ -30 = 10a
➙ a = -3 m/s²
If the speed is decreasing with time then accerlation is negative. The negative accerlation is called deceleration or retardation.
Thus, the retardation of the bus is 3 m/s².
Now, using 2nd equation of motion that is,
s = ut + 1/2at²
By substituting the values in the equation,
➙ s = ut + 1/2at²
➙ s = (30)(10) + 1/2(- 3)(10)²
➙ s = 300 - 150
➙ s = 150 m
Thus, the distance travelled by the bus is 150m.
Bonjour ⚘⚘
Answer:
150 m
Explanation:
Given information : A bus moving at a speed of 30ms^-1 is brought to rest within 10s by applying.
by using 1st equation of motion:
v = u + at
where,
- v : final velocity
- u : initial velocity
- a : acceleration
- t : time taken
>> v = u + at
>> 0 = 30 + a*10
>> -30 = 10a
>> a = -3 m/s^2
now finding the distance by using 2nd equation of motion
s = ut + 1/2at^2
where,
- s : distance travelled
- u : initial velocity
- t : time taken
- a : acceleration
- t : time taken
>> s = ut + 1/2at^2
>> s = (30)(10) + 1/2(- 3)(10)^2
>> s = 300 -150
>> s = 150 m
distance travelled by the bus is 150m.