How to calculate for the complete trip A record of travel along a straight path

Question

How to calculate for the complete trip A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.50 m/s2 for 11.0 s 2. Maintain a constant velocity for the next 2.90 min. 3. Apply a constant negative acceleration of -9.48 m/s2 for 2.90 s. (a) What was the total displacement for the trip? (b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip? leg 1 leg 2 leg 3 complete trip m/s m/s m/s m/s Need Help? Read lt SerCP11 2.P040 triver then applies the bra

Explanation / Answer

Given,

a = 2.5 m/s^2 for t = 11 s ; t1 = 2.9 min = 174 ; a1 = -9.48 m/s^2 for t1 = 2.9 s

a)The displacement in the first leg:

x1 = 1/2 at^2 = 0.5 x 2.5 x 11^2 = 151.25 m

velocity at the end of first leg:

v1 = at = 2.5 x 11 = 27.5 m/s

displacement will be:

x2 = v t = 27.5 x 174 = 4785 m

speed remains same,

distance moved while it slows:

x3 = 1/2 at^2 = 0.5 x 9.48 x 2.9^2 = 39.86 m

Displacement is:

X = x1 + x2 + x3 = 151.25 + 4785 + 39.86 = 4976.11 m

Hence, X = 4976.11 m

Leg 1:

Vav = (v0 + v1)/2 = 27.5/2 = 13.75 m/s

Leg 2

Vav = 27.5 m/s

Leg 3

Vav = (v0 + v1)/2 = 27.5/2 = 13.75 m/s

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