Kinematics 5 Question 3

5. The position vector of particle changes with time according to the relation r(t)=15t2i^+(420t2)j^. What is the magnitude of the acceleration (in ms2 ) at t=1 ?

(a) 50

(b) 100

(c) 25

(d) 40

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Answer:

Correct Answer: 5. (a)

Solution:

  1. Position vector of particle is given as

r=15t2i^+(420t2)j^

Velocity of particle is

v=drdt=ddt[15t2i^+(420t2)j^]=30ti^40tj^

Acceleration of particle is

a=ddt(v)=ddt(30ti^40j^)=30i^40j^

So, magnitude of acceleration at t=1s is

|a|t=1s=ax2+ay2=302+402=50ms2



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