Question

10. An object initially at rest experiences an acceleration of 1.20 m/s2 for 6.10 s then travels at that constant velocity for another 8.40 s. What is the magnitude of the objects average velocity over the 14.5 s interval? m/s

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Answer #1

Given is:-

acceleration a-1.20m /s

t_1 = 6.10s, t_2 = 8.40s

Now,

the final velocity of the object will be

v = u +at_1

by plugging all the values we get

v-0 (1.20 (6.10)

which gives us

v7.32m/s

distance travelled during this time interval is

s_1 = ut + rac{1}{2}at_1^2

s1 (0)(6.10) +-(1.20)(6.10)2

si 22.326772

Now,

the distance travelled during time t2 is

s_2 = v imes t_2

89 7.32 × 8.40

s261.488m

total distance travelled by the object is

s = s_1 + s_2

s 22.326 61.488

s83.814m

and total time taken is

t t1 + t2 = 14.5s

Now,

the average velocity is given by

aug tf-ti

by plugging all the values we get

avg.V83.814 -0 14.5-0 aug

which gives us

avg.V 5.78m/s

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