Starting from a stationary position anil paddles hia bicycle to attain a velocity of 10m/s in 25 seconds then he applies breaks,such that he again comes to rest after 50 seconds.Calculate the acceleration of the bicycle in both cases and also find the total distance covered by anil.
Answers
In 1st case the acceleration is 0.4 m/s^2 and distance covered is 125 m.
In 2nd case the acceleration is -0.2 m/s^2 and distance covered is 750 m.
Explanation:
CASE 1 :
initial velocity = 0 (u)
final velocity = 10 m/s (v)
time = 25 s (t)
- acceleration = ?
- distance = ?
a = v - u/t
a = 10-0/25
a =10/25
a =0.4 m/s^2
- From 2nd equation of motion
s = ut +1/2 . at^2
s = 0 x 25+1/2 x 0.4 x (25) ^2
s = 0+125
s = 125 m
CASE 2 :
from above we know that
initial velocity = 10 m/s (u)
final velocity = 0 (v)
time = 50 s (t)
acceleration = ?
a = v-u/t
a = 0-10/50
a = -10/50
a = -0.2 m/s^2
- From 2nd equation of motion
distance = ?
a = -0.2 m/s square
t = 50 s
u = 10 m/s
- Distance covered:
s = ut+1/2 at square
s=10 x 50 + 1/2 x -0.2 x (50)^2
s = 500+250
s = 750 m
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Answer:
CASE 1
initial velo.= 0 (u)
final velo.= 10m/s (v)
time = 25s (t)
acceleration = ?
distance = ?
a = v-u/t
a = 10-0/25
a =10/25
a =0.4m/s(square)
now from 2nd equation of motion
s = ut +1/2at(square)
s = 0X25+1/2X0.4X25square
s = 0+125
s = 125m
CASE 2
from above we know that
initial velo. = 10m/s (u)
final velo. = 0 (v)
time = 50s (t)
accelaration = ?
a = v-u/t
a = 0-10/50
a = -10/50
a = -0.2m/s (square)
now from 2nd equation of motion
distance = ?
a = -0.2m/s square
t = 50s
u = 10m/s
s = ut+1/2 at square
s=10X50+1/2X-0.2X50square
s = 500+250
s = 750m