find the initial velocity of a car which is stopped in 10 sec by appling breaks ? The retardation due to breaks is 2 .5m/s
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Answered by
3
hey there
equation is
v =u + at
it comes to rest so v=0
0 = u + 2.5× 10
u = -25 m /s
negative sign denotes initial velocity was in opposite direction of retardation
equation is
v =u + at
it comes to rest so v=0
0 = u + 2.5× 10
u = -25 m /s
negative sign denotes initial velocity was in opposite direction of retardation
ATUL2411:
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u= -25m/s is wrong
no it correct
u=25m/s is right
no you took dirn of u positive and i took acc to be positive
but acceleration is in retardation so acceleration is negative
no i took the initial velocity to be negtive so accn became poaitive
but a equal to v minus u by t when we put u equal to negative in then u will be positive
Answered by
1
a= v-u/t
a= -2.5m/s^2
v=0
t=10 sec
-2.5=0-u/10
-u=-2.5×10
-u=-25
u=25
a= -2.5m/s^2
v=0
t=10 sec
-2.5=0-u/10
-u=-2.5×10
-u=-25
u=25
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