Posted by **Anne** on Sunday, November 11, 2012 at 3:26pm.

A certain car is capable of accelerating at a uniform rate of 0.93 m/s2. What is the magnitude of the carâ€™s displacement as it accelerates uniformly from a speed of 86 km/h to one of 98 km/h?

Answer in units of m

- physics -
**bobpursley**, Sunday, November 11, 2012 at 3:55pm
vf^2=vi^2+2ad solve for d.

- physics -
**bobpursley**, Sunday, November 11, 2012 at 3:55pm
be certain to change km/hr to m/s...

- physics -
**Damon**, Sunday, November 11, 2012 at 4:04pm
v = Vi + a t

Vi = 86 km/h * 1000 m/km * 1 h/3600 s

= 23.9 m/s

Vf = 98 km/h *10/36 = 27.2 m/s

hard way:

27.2 = 23.9 + .93 t

so

t = 3.55 s

x = Xi + Vi t + (1/2) a t^2

x-Xi = displacement = 23.9(3.55)+.465(3.55)^2

= 90.7 m

easy way:

average speed = (23.9+27.2)/2 = 25.55

change in speed/t = .93 = 3.3/t

t = 3.55 sure enough

distance = 25.55*3.55 = 90.7 m

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