a car is moving with a velocity of 12 M/S. on applying brakes, it comes to rest in 6 second. calculate the (1) acceleration (2) distance covered in the time period.
Answers
Answered by
7
Hi there !!
Here's your answer :)
Given that ,
The car is moving with a velocity of 12m/s
Initial velocity (u) = 12m/s
Final velocity (v) = 0m/s [ the car comes to rest ]
Time (t) = 6 sec
Let the acceleration be a
To find the acceleration ,
We know,
a = v-u/t
= 0 - 12/6 = -12/6 = -2
Thus,
The acceleration is 2m/s² in the direction opposite to that of motion.
To find the distance , we will use the 3rd equation of motion
That is,
2as = v² - u²
2(-2)(s) = 0² - 12²
-4s = -144
4s = 144
s = 36
Thus,
The distance is 36 m
________________
Hope it helps :D
Here's your answer :)
Given that ,
The car is moving with a velocity of 12m/s
Initial velocity (u) = 12m/s
Final velocity (v) = 0m/s [ the car comes to rest ]
Time (t) = 6 sec
Let the acceleration be a
To find the acceleration ,
We know,
a = v-u/t
= 0 - 12/6 = -12/6 = -2
Thus,
The acceleration is 2m/s² in the direction opposite to that of motion.
To find the distance , we will use the 3rd equation of motion
That is,
2as = v² - u²
2(-2)(s) = 0² - 12²
-4s = -144
4s = 144
s = 36
Thus,
The distance is 36 m
________________
Hope it helps :D
Anonymous:
^_^
Similar questions