Applications Of Derivatives Question 38
Question: The displacement of a particle in time t is given by $ s=2t^{2}-3t+1 $ . The acceleration is
Options:
A) 1
B) 3
C) 4
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
$ s=2t^{2}-3t+1;,\therefore \frac{ds}{dt}=v=4t-3 $ At $ t=0{{( \frac{ds}{dt} )}{t.0}}=-3=v_1 $ Now, at $ t=1 $ , we get $ {{( \frac{ds}{dt} )}{t=1}}=4-3=1=v_2 $
Hence rate of velocity $ =v_2-v_1=1-(-3)=4 $ Aliter : Given that $ s=2t^{2}-3t+1 $
$ \frac{ds}{dt}=4t-3 $ (velocity), Again $ \frac{d^{2}s}{dt^{2}}=4 $ (acceleration).
BETA
