Question

A car is moving with a velocity of 10 m/s.
Suddenly driver sees a child at a distance of 30 m
and he applies the brakes, it produced a uniform
retardation in car and car stops in 5 second.
Find (mass of car=1000 kg)
(i) the stopping distance of car
(ii) will the car hit the child
(iii) the retardation in car
(iv) resistive force on car​

Answer 1

Answer:

look above at the picture

Answer 2

Answer:-

S = 25 m

No

a = -2m/s²

F = -2000 N

Given :-

u = 10 m/s

Distance of child = 30 m

v = 0m/s

t = 5 s

To find :-

(i) the stopping distance of car

(ii) will the car hit the child

(iii) the retardation in car

(iv) resistive force on car

Solution:-

Since,the driver applied brakes and car retarted uniformly.

So , let's find the uniform retardation of car :-

$\huge \boxed {a = \dfrac{v-u}{t}}$

→$a = \dfrac{0-10}{5}$

→$a = \dfrac{-10}{5}$

→$a = -2 m/s^2$

Hence, the uniform retardation of car will be -2 m/s².

Now, find the distance travelled by car after applying the brakes.

$\huge \boxed {2as = v^2 - u^2}$

→$2 \times -2 \times s = (0)^2 - (10)^2$

→$-4\times s = -100$

→$s = \dfrac{-100}{-4}$

→$s = 25 m$

Since,the driver applied brakes at 25 m but the child is at 30 m .

the driver applied brakes at 25 m but the child is at 30 m .hence, it stops before 5 m and the child is saved from accident.

We have to find the resistive frictional force in the car tyres.

m = 1000 kg

a = -2 m/s²

From Newton 2nd law of motion:-

$\huge \boxed {F = ma}$

→$F = 1000 \times -2$

→$F = -2000 N$

The negative sign shows that force applied is opposite in direction of car.

hence, the magnitude of resistive force will be -2000 N.