the brakes are applied to a car produce on acceleration of 6m/s in the opposite direction of the motion . If the car takes 2s to stop after the application of brakes . Calculate the distance traveled by the car during the time
Answers
Given that the acceleration i
was produced in the opposite direction of the motion. So it is retardation.
a = -6m/s²
Final velocity of the car = 0
Time = t = 2s
So finding initial velocity of the car by the relation ,
v = u + at
0 = u +(-6)2 = u-12
u = 12 m/s
Then, finding the distance covered during that time , by using the relation : .
s = ut + ½ at²
= 12× 2+½+(-6)×2² = 24 - 12
= 12 m
The required distance = 12 m
GIVEN::
Acceleration= -6m/s² [Since, it is retardation]
Final velocity (v) = 0 [Since, the car stops finally.]
Time taken = 2 sec
Initial velocity(u) = ?
Distance travelled(s) = ?
LETS FIND INITIAL VELOCITY (u)
To find velocity we have to apply first equation of motion i.e. v=u + at
⇒v = u + at
⇒0 = u + (-6 × 2)
⇒0 = u - 12
⇒-u = -12
⇒u = 12.
Therefore, initial velocity is 12m/s.
LETS FIND THE DISTANCE TRAVELLED (s)
To find the distance travelled we have to apply 2nd equation of motion.
⇒s = ut + 1/2at²
⇒s = (12 × 2) + (1/2 × -6 × 2²)
⇒s = 24 + (-3 × 4)
⇒s = 24 - 12
⇒s = 12m.
Therfore, the distance travelled by the car is 12m.