Question

A car going at a speed of 7ms^-1 can be stopped by applying brakes in a shortest distance of 10m . Show that the total frictional force opposing the motion .when brakes are applied is 1/4th of the weight of the car . Answer step by step ...​

Answer 1

For a motor car:

u=7m/s; v=0; s=10m

Using v

2

=u

2

+2as, we get

a=

2s

v

2

−u

2

=

2×10

0−7

2

=−2.45m/s

2

a=−g/4

−f=−ma=mg/4

1/4

th

of the weight of the car.

Answer 2

Hint: By seeing the question we can say that if the car stopped its final velocity will become 0, Hence by given data we can calculate acceleration. Also the mass of the car is not given in the question but we know that force is mass multiplied by product. From force we can find mass.

Formula used:

(i)V2 = u2 + 2as

V

is final velocity

u

is initial velocity

a

is acceleration

(ii) Force  =  Mass×acceleration

F = m×a.

Complete step by step answer:

We have given, a car is at speed 7ms - 1 moving and suddenly it stopped by applying brakes so its final velocity become 0. and distance travelled by it is 10m. We have to show the total resistance force (functional force) is 14 times the weight of the car.

Data given,

u = 7m/s,V = 0,s = 10m,& acceleration = ?

So,

By using 3rd equation of Motion

V = u2 + 2as

So, a = V2 - u22s

a = (0)2 - (7)22×10

⇒a =  - 4920⇒ - 2.45m/s2

Now,

2.45 can also be written as a=4920

⇒a=49×220×2=9840=9.84=g4(∵g=9.8)

Now, the resistance or frictional force  = ma

So, the total frictional force opposing the motion in 14 times the weight of car R(functional force) =14 weight of car.

Note: In order to solve this question, we need to count the acceleration in terms of g as you have seen in the solution. Negative sign shows that the motion is in the opposite direction with the frictional force or resistance force. It must be noted that frictional force is always in the opposite direction of the motion.