The position vector of a particle change with time according to the relation 
(t)=15t2
+(4−20t2)
. What is the magnitude of the accelerating at t=1?
(a)
50
(b)
100
(c)
40
(d)
25
The position vector of a particle change with time according to the relation (t)=15t2 +(4−20t2) . What is the magnitude of the accelerating at t=1?
(a)
50
(b)
100
(c)
40
(d)
25
Question ID - 51269 | SaraNextGen Top Answer
The position vector of a particle change with time according to the relation (t)=15t2 +(4−20t2) . What is the magnitude of the accelerating at t=1?
(a)
50
(b)
100
(c)
40
(d)
25
|
The position vector of a particle change with time according to the relation |
|||
|
(a) |
50 |
(b) |
100 |
|
(c) |
40 |
(d) |
25 |
1 Answer
127 votes
=15t2
+4
20t2
=30t
−40t
=30
−40
=50m/s2127