Question

can someone explain it with steps?

A car was moving on a road at a constant speed of 15 m/s when suddenly the car driver saw some animal on the road at a distance of 21 m from the car, so he applied the brakes after a response time of 0.4 s and stopped before hitting the animal by 1 m. What was the deceleration of the car?

a-7.5 m/s^2

b-5.2 m/s^2

c-8.0 m/s^2

d-5.6 m/s^2​

Answer 1

Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

$=15\times0.4\\\\=6m$

The car stopped before hitting the animal by $1 m$

Distance travelled during deceleration is $21-6-1=14m$

Hence by $v^2=u^2+2as$

We have

$0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2$

Option C is the correct answer

Answer 2

• Initial velocity=15m/s=u
• Final velocity=0
• deacceleration=a

Distance traveled during reaction time

• 15(0.4)=6m

Total distance

• 21-6-1=14m

$\\ \sf\longmapsto v^2-u^2=2as$

$\\ \sf\longmapsto -(15)^2=2(14)a$

$\\ \sf\longmapsto -225=28a$

$\\ \sf\longmapsto a=-8.0m/s^2$