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Question : 178 of 180
Marks:
+1,
-0
Solution:
We have,
S2=at2+2bt+c . . . (i)
Differentiating Eq. (i) w.r.t.
t, we get
2S=2at+2b⇒=() . . . (ii)
Differentiating Eq. (ii) w.r.t.
t, we get
=⇒= [using Eq. (ii)]
⇒=⇒=| a(at2+2bt+c)−[(at)2+b2+2atb] |
| S3 |
[using Eq. (i)]
⇒=| a2t2+2abt+ac−a2t2−b2−2atb |
| S3 |
⇒=⇒ Acceleration =()∝
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