A motor car is uniformly retarded and brought to rest from velocity 36km/hours in 5 second period it is retardation and distance covered during this period
Yes but i want to solv pls help me solv
To find the retardation and distance covered during the period, we can use the equations of motion.
1. Convert the initial velocity from km/hr to m/s:
Initial velocity (v0) = 36 km/hr = (36 * 1000) m/3600s = 10 m/s
2. Final velocity (v) is 0 m/s as the car comes to rest.
3. Time taken (t) = 5 seconds
We can use the equation of motion:
v = v0 + at
where,
v = final velocity (0 m/s)
v0 = initial velocity (10 m/s)
a = acceleration/retardation
t = time (5 seconds)
Rearranging the equation, we get:
a = (v - v0) / t
Substituting the values into the equation:
a = (0 - 10) / 5
= -10 / 5
= -2 m/s²
The negative sign indicates that the car is experiencing retardation.
Therefore, the retardation (a) of the car is -2 m/s².
Now, to find the distance covered (s) during the 5-second period, we can use another equation of motion:
s = v0t + (1/2)at²
where s = distance covered
Substituting the values:
s = 10 * 5 + (1/2) * (-2) * (5²)
= 50 - 10 * 5
= 50 - 50
= 0 m
Therefore, the distance covered (s) during the 5-second period is 0 meters.
Can I see the full explanation.
Yes
a = -36km/hr / 5s = -7.2 km/hr/s
distance s = vt + 1/2 at^2, and you know t=5