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Question:

The position vector of a particle changes with time according to the relation r(t)=15t²i^+(4-20t²)j^. What is the magnitude of the acceleration at t=1?

25

40

100

50

Solution:

Correct option is D. 50
r→(t)=15t²i^+(4−20t²)j^
r→'(t)=15(2t)i^+(4−20(2t))j^
r→'(t)=30ti^+(0−40t)j^
r→'(t)=30ti^−40tj^|r→'(t)|=(30)²+(−40)²=900+1600=2500=50