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`        A particle starts from rest with a fconstant acceleration. At a time t second, the speed is found to be 100m/s and one second later the speed becomes 150m/s. find (a) the acceleration and (b) the distance travelled during the (t+1)th second.`
8 years ago

Sachin Tyagi
31 Points
```										(a) Velocity at time t is 100m/s =a.(t second).............................1
and velocity at time (t+1) second is  150m/s=a. (t+1) second....................2
(b) consider the interval t second to (t+1) second, time elapsed =1s
Initial Velocity = 100 m/s
Final velocity = 150 m/s.
Thus,  (150 m/s)2 = (100m/s)2 + 2(50 m/s2 )x
or,                        x= 125 m.
```
8 years ago
bala krishna
33 Points
```										This is a simply question that can be solved in more than one method.

Given,

Let velocity at the instant time 't' = 100 m/s
now velocity after 1 second that is at 't+1' sec velocity =150

Let acceleration be a.

now from equations of motion,
v = u +at
150 = 100 + at' ...............1

where t' = [t+1]-t =1 sec.

there fore a = 50 m/s2.      [a bit]

distance travelled :

v2 -u2 = 2as

there fore,2500 m
```
8 years ago
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