Question

A sports car is advertised to be able to stop in a distance of 45 m from a speed of 87 km/hr. what is its acceleration? (a) What is its acceleration in m/s^2? (b) How many g's is this (g = 9.80 m/s^2)?

Answer 1

Answer:
`6.49 m/s^2`

Step-by-step explanation:

Given

distance =45 m

initial velocity(u)
`=87 km/hr\approx 24.1667 m/s`

Final velocity(v)=0

`v^2-u^2=2as`

`0-(24.1667)^2=2* a* 45`

`a=(584.029)/(2* 45)`

`a=-6.49 m/s^2`

acceleration in terms of g

`(a)/(g)=0.662`

acceleration=0.662g

Answer 2

a) -6.48 m/s²

b) a = 0.661g

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(0^2-\left((87)/(3.6)\right)^2)/(2* 45)\\\Rightarrow a=-6.48\ m/s^2`

Acceleration of the car is -6.48 m/s²

g = 9.81 m/s²

`(a)/(g)=(6.48)/(9.81)\\\Rightarrow (a)/(g)=0.661\\\Rightarrow a=0.661g`

a = 0.661g