The maximum deceleration of most cars is about 7m/s. Find the braking distances at 14MIS,21m/s and 28m/s.?

Question

u=initial speed(speed at the start)

v=final speed

a=acceleration

t=time

s=displacement(distance travelled)

formulas:

v= u + at

s= ut + 1/2atsquared

vsquared=usquared + 2as

you can use them if u want but u can do ur own way...thankyou in advance..i nid it asap..pls2..help!..

Answer 1

I assume you mean 7 m/s^2

0=21^2 -2x7xdistance as final speed is 0

441=2x7xdistance

distance = 31.5 m

0=28^2-2x7xdistance

distance = 56m

I am unfamiliar with MIS

Answer 2

You want to find the braking distance, S. You know acceleration (a), start velocity (u) and final velocity (v). The only equation that will do is:

vsquared=usquared + 2as

2as = (v^2) - (u^2)

s= ((v^2) - (u^2)) / 2a

a=-7m/s/s

v = 0 m/s (ie. stopped)

if u=14m/s

s= ((0 * 0) - (14 * 14)) / (2 * -7)

s= -196 / -14

s=14 m

You should now be able to get the answer for 21m/s and 28 m/s