Please help me question 2.41 & question 2.44 in Physic 139. The driver of a pick

Question

Please help me question 2.41 & question 2.44 in Physic 139.
The driver of a pickup truck going 100 km/h applies the brakes, giving the truck a uniform deceleration of 6.5 m/s^2 while it travels 20.0 m. (a) What is the speed of the truck in kilometers per hour at the end of this distance? (b) How much time has elapsed? Two identical cars capable of accelerating at 3.00 m/s^2 are racing on a straight track with running starts. Car A has an initial speed of 2.5 m/s: car B starts with speed of 5.00 m/s. (a) What is the separation of the two cars after 10 s? (b) Which car is moving faster after 10 s?

Explanation / Answer

2.41

Vi = 100 km/hr = 27.78 m/sec

a = -6.5 m/sec^2

s = 20 m

Using equation:

Vf^2 = Vi^2 + 2*a*s

Vf = sqrt (27.78^2 - 2*6.5*20) = 22.62 m/sec

Vf = 22.62*(18/5) = 81.43 km/hr

B.

vf = vi + a*t

t = (vf - vi)/a

t = (22.62 - 27.78)/(-6.5)

t = 0.794 sec

2.44

V1i = 2.5 m/sec, V2i = 5 m/sec

a1i = a2i = 3 m/sec^2

after 10 sec,

s = u*t + 0.5*a*t^2

s1 = 2.5*10 + 0.5*3*10^2 = 175 m

s2 = 5*10 + 0.5*3*10^2 = 200 m

separation = 200 - 175 = 25 m

B.

V1f = V1i + a*t

V1f = 2.5 + 3*10 = 32.5 m/sec

V2f = 5 + 3*10 = 35 m/sec

Car B is running faster

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