Question

The brakes applied to a car produces deceleration of 10 ms-2 in opposite direction to the motion. If car requires 2.5 sec. to stop after application of brakes, calculate distance travelled by the car during this time​

Answer 1

Answer:

Given:

Accerleration: a=−6 m/s

2

Time t=2 s

Final velocity, v=0 m/s

v=u+at

0=u−6×2

u=12 m/s

s=ut+

2

1

at

2

s=12×2+

2

1

×(−6)×4

⇒s=12 m

Hope it helps you!

Answer 2

Answer:

Distance = 31.25 m

Explanation:

Given -

• Retardation/declaration(a) = - 10m/s²
• Final velocity (v) = 0m/s
• Time (t) = 2.5 seconds

Solution -

Finding initial velocity (u) -

• v=u+at
• u= at-v
• u= (2.5×10)
• u= 25m/s

Initial velocity is 25m/s.

Finding distance(s) -

• s= ut+1/2at²
• (0×2.5)+1/2×10×2.5²
• 5×6.25
• 31.25

Distance is 31.25m .