a motor car is uniformly retarded and brought to rest from a velocity of 50km/h in 10sec. find its retardation and the distance covered during this period?

the acceleration is clearly (-50km/hr)/(10s) = -5 km/hr/s

now, 1 km/hr = 5/18 m/s
so, a = -25/18 m/s^2

now to find the distance,
s = 250/18 t - 25/36 t^2
plug in t=10

No

It is correct

To find the car's retardation and the distance covered during the period, we can use the following equations of motion:

1. v = u + at
2. v^2 = u^2 + 2as

where:
- v is the final velocity (0 km/h since the car comes to rest)
- u is the initial velocity (50 km/h)
- a is the retardation (what we need to find)
- t is the time (10 seconds)
- s is the distance covered (what we need to find)

First, let's find the retardation:
Using equation 1, we can rearrange it to solve for a:

v = u + at
0 = 50 + a * 10

Simplifying:
-50 = 10a
a = -5 km/h^2 (the negative sign indicates retardation)

The retardation is -5 km/h^2.

Next, let's find the distance covered:
Using equation 2, we can rearrange it to solve for s:

v^2 = u^2 + 2as
0 = (50)^2 + 2 * (-5) * s

Simplifying:
0 = 2500 - 10s
10s = 2500
s = 250 km

The distance covered during this period is 250 km.

Therefore, the retardation is -5 km/h^2 and the distance covered is 250 km.

I did,not understand

Saminu

Vo = 50km/h = 50,000m/3600s = 13.9 m/s.

V = Vo + a*t = 0.
13.9 + a*10 = 0,
a = -1.39 m/s^2.

V^2 = Vo^2 + 2a*d = 0.
13.9^2 + 2*(-1.39)*d = 0,
d = 69.5 m.