Question

the brakes of an automobile can produce a constant deceleration of 3 meter per second square . find the maximum distance travelled after the application of brakes if it comes to rest in 6 sec . ANS. 54​

Answer 1

ANSWER

54 m

CONCEPT

• KINEMATIC EQUATIONS
• DERIVATION GIVEN IN NCERT

FORMULA USED

V = U + at

s = ut +1/2at^2

EXPLANATION

since automobile deacelerate so after some time it comes at rest so final velocity v = 0

given time = t = 6 sec

a = - 3 since - sign indicated that car is deacelerate Or we can say acceleration in opposite direction

applying formula

0 = U -3×6

we get u = 18 m per sec

now ,

s = ut +1/2at^2

= 18×6 -1/2×3×6×6

=54 m

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Answer 2

Answer:

The distance travelled by the automobile is 54m

Explanation:

Given ,

The deceleration of the automobile , a = -3m/s²

The time taken to stop the automobile , t = 6s

We have , final velocity = 0m/s ( since the body stops )

initial velocity , u = ?

Now ,

$v = u + at \\ \implies0 = u + ( - 3) \times 6 \\ \implies \: u = 18m {s}^{ - 1}$

Again we have from equations of motion

$s = ut + \frac{1}{2} a {t}^{2} \\ \implies s = 18 \times 6 + \frac{1}{2} \times( - 3 )\times 6 ^{2} \\ \implies s \: = 108 - 54 \\ \implies s \: = 54m$