A car starting out at a velocity of 45km/hr, experiences an acceleration of a =

Question

A car starting out at a velocity of 45km/hr, experiences an acceleration of a = -0.50m/s^2, which slows it down till it comes to rest. a. Determine the distance traveled. b. Calculate the time it takes to come to rest. c. Now does x change with time between t=7.0s and t=8.0s? Two cars start at the same position and time: The green car initially moves at a velocity= 8.00m/s and has an acceleration of a=-0.600m/s^2, and The red car has an initial velocity of 3.00m/s and has an acceleration of a=+0.600m/s2. When does the red car catch up to the green car? Hint: When they have the same position of have traveled the same distance]

Explanation / Answer

u = 45 = 45*5/18 = 12.5 m/s

a)

v^2 - u^2 = 2as

0 - 12.5^2 = 2*-0.5*s

s= 156.25m

b) v = u +at

0 = 12.5 - 0.5*t

t = 25s

c) x1 = ut1 + 0.5*a*t^2 = 12.5*7 - 0.5*0.5*49 = 75.25 m

x2 = ut2 + 0.5*a*t^2 = 12.5*8 - 0.5*0.5*64 = 84 m

sR = sG

(8*t) - (0.5*0.6*t^2) = (3*t)+(0.5*0.6*t^2)

8t - 0.3*t^2 = 3t + 0.3*t^2

5t = 0.6t^2

t = 8.3 s

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