Question

a car, running at a speed of 72km/h is slowed down to 18km/h over a distance of 40m. calculate the retardation produced by its brakes, and time for which brakes are applied​

Answer 1

answEr -

u = 72km/h

v = 18km/h

s= 40m

s= ut -1/2 at²

or,

v² = u²- 2as

as we got acceleration= - 375/80

or, a= -4.6875m/s²

now,

we know

a= (v-u)/t

-375/80 = (5-20)/t

or,

t= -15x-375/80

or,

t= 3.2 sec

done

Answer 2

Explanation:

We have given;

Initial velocity(u) = 72km/h

In m/s, initial velocity(u) = 20m/s;

Final velocity(v) = 18km/h

In m/s, final velocity(v) = 5m/s;

Distance travelled = 40m;

(here, we will use all the units in meter/seconds and meter. We can even use the unit in km/hr and km.)

Formula to be applied :

v^2 = u^2 + 2as;

(we are using this formula because we have given v, u and s and has asked for a.)

So, put all the given values in the formula:

(5)^2 = (20)^2 + 2(a)(40);

25 = 400 + 80a;

-375 = 80a;

a = -375/80;

= -4.6875 m/s^2.

Here, retardation will be 4.6875 m/s^2.

-4.6875 m/s^2 is the acceleration.

Now, as for the time, the formula to be applied will be;

v = u + at;

Thus, put the values in this formula;

5 = 20 + (-375/80)(t)

-15 = (-375/80)(t)

t = (-15)(80/-375)

= 3.2 seconds.

That's all.