Question

the brakes applied to a car produce an acceleration of 6m/s^2 in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time.

Answer 1

Heya friend,
Here is the solution:

Initial velocity(u) = ?
Final velocity (v) = 0 m/s, as the object stops after the brakes are applied.
Retardation (a)   = - 6 m/$s^{2}$ , retardation is negative acceleration
Distance (s)        = ?
Time (t)               = 2 seconds


We know,
From first equation of motion, i.e. v = u + at

⇒ 0 = u + -6 × 2
⇒ - u = -12
⇒ u = 12 m/s

Now, from third equation of motion, i.e. s = ut + 1/2 a$t^{2}$
We get,

⇒ s = 12 × 2 + 1/2 (-6 ) × ()$2^{2}$
⇒ s = 24 + 1/2 × (-6 × 4)
⇒ s = 24 + 1/2 (-24)
⇒ s = 24 - 12
⇒ s = 12 meters(m)

∴ So, the distance traveled by car after brakes been applied is 12 m .



Thank You!!!
@MANAV