Question

Determine the stopping distances for an automobile going a constant initial speed of 95 km/h and human reaction time of 0.40 s: a) for an acceleration a = -3.0 m/s2; b) for a = -6.0 m/s2.

Answer 1

Answer:

as given

initial velocity, u = 95kmph= 6.08 m per sec

time =0.40 s

a1= -3 m/s²

a2= -6m/s²

now

S= ut+ 1/2a1t²

so, S1= 6.08× 0.40 + 1/2 ×(3 ) ×(0.40)²

= 1.456m

similarly S2= it +1/2 a2t²

= 6.08×0.40 +1/2 (6)×( 0.40)²

=1.696 m

Explanation:

NOTE : NEGATIVE ACCELERATION IS TAKEN AS RETARDATION WHILE FINDING DISTANCE TRAVELLED AND TIME TAKEN

THE MAIN DIFFERENCE IS FOR EX:

Acceleration -4m/s²= retardation 4m/s²

bo negative sign in retardation as it is negative of acceleration

Answer 2

Given: An automobile going a constant initial speed of 95 km/h and human reaction time of 0.40s.

To find: The stopping distances for an acceleration,

(a) a = -3.0 ms⁻²

(b) a = -6.0 ms⁻².

Solution:

• According to the distance-time equation, we have,

       $S = ut + \frac{1}{2} at^{2}$

• Here, S is the displacement, u is the initial velocity of the automobile, t is the time taken for the automobile to stop and a is the acceleration produced.
• The initial velocity given is 95kmh⁻¹, that is equal to 26.39 ms⁻¹.

(a)

• Here, the acceleration is -3.0 ms⁻².
• So, according to the formula,

       $S = (26.39ms^{-1} )(0.4s) + \frac{1}{2} (-3.0ms^{-2})(0.4s)^{2}$

       $S = 10.316 m$.

(b)

• Here, the acceleration is -6.0 ms⁻².
• So, according to the formula,

        $S = (26.39ms^{-1} )(0.4s) + \frac{1}{2} (-6.0ms^{-2})(0.4s)^{2}$

        $S = 10.076m$.

Therefore, the stopping distances for an automobile

(a) for a = -3.0 ms⁻² is 10.316 m.

(b) for a = -6.0 ms⁻² is 10.076 m.