a body moves with a uniform velocity of 2m/sec for 5 sec then velocity was uniformly accelerated and a velocity of 10m/sec was achieved in next 5 sec.then brakes were applied uniformly and body comes to rest in next 10 sec.calculate
a)acceleration and retardation produced
b)distance moved by the body with uniform velocity
c)different distances moved by the body with non-uniform velocity i.e.when acceleration was active
d)average velocity in whole journey
s1=v1•t1 =2•5 =10 m.
v2=v1+a1•t2 .
a1 =(v2-v1)/t2 =( 10-5)/5 = 1m/s².
s2 = v1•t2 + a1•t2²/2 =5•5+1•25/2 =37.5 m.
0 =v2+a2•t3,
a2 =(0-v2)/t3 = -10/10 =-1 m/s².
s3 = v2•t3 – a2•t3²/2 =10•10-1•100/2 =50 m
v(ave_ =s/t =(s1+s2+s3)/(t1+t2+t3) =
(10+37.5+50)/(5+5+10) = 97.5/20 =4.875 m/s
To solve this problem, let's break it down into different parts:
a) Calculation of acceleration and retardation:
We know that acceleration is the change in velocity divided by the time taken, and retardation (or deceleration) is the negative value of acceleration.
Given:
Initial velocity (u) = 2 m/sec
Final velocity (v) = 10 m/sec
Time taken to achieve the final velocity (t1) = 5 sec
Time taken to come to rest (t2) = 10 sec
Acceleration (a) can be calculated using the equation:
a = (v - u) / t1
a = (10 - 2) / 5
a = 8 / 5
a = 1.6 m/sec^2
Retardation (r) can be calculated using the equation:
r = -u / t2
r = -2 / 10
r = -0.2 m/sec^2
b) Calculation of distance moved with uniform velocity:
Distance (S1) moved with uniform velocity can be calculated using the equation:
S1 = u * t1
S1 = 2 * 5
S1 = 10 m
c) Calculation of different distances moved with non-uniform velocity:
To calculate the different distances moved with non-uniform velocity, we need to find the distance covered during acceleration and retardation separately.
Distance covered during acceleration (S2) can be calculated using the equation:
S2 = (u * t + (1/2) * a * t^2)
S2 = (2 * 5 + (1/2) * 1.6 * 5^2)
S2 = 10 + (1/2) * 1.6 * 25
S2 = 10 + 20
S2 = 30 m
Distance covered during retardation (S3) can be calculated using the equation:
S3 = (v * t + (1/2) * r * t^2)
S3 = (10 * 10 + (1/2) * (-0.2) * 10^2)
S3 = 100 + (-0.1) * 100
S3 = 100 - 10
S3 = 90 m
d) Calculation of average velocity in the whole journey:
The average velocity (Vavg) can be calculated by summing up the total distances and dividing it by the total time taken:
Vavg = (S1 + S2 + S3) / (t1 + t1 + t2)
Vavg = (10 + 30 + 90) / (5 + 5 + 10)
Vavg = 130 / 20
Vavg = 6.5 m/sec
So, the answers are:
a) The acceleration produced is 1.6 m/sec^2 and the retardation produced is -0.2 m/sec^2.
b) The distance moved by the body with uniform velocity is 10 meters.
c) The different distances moved by the body with non-uniform velocity are 30 meters during acceleration and 90 meters during retardation.
d) The average velocity in the whole journey is 6.5 m/sec.