Question

Help please

Answer 1

A car comes to a stop from a speed of 30ms^-1 in a distance of 804m. The driver brakes so as to produce a deceleration of 1/2ms^-2 to begin with, and then brakes harder to produce a deceleration of 3/2ms^-2. Find the speed of the car the instant when the deceleration is increased and the total time the car takes to stop.

Answer 2

this is physics

Answer 3

@KingGeorge @AccessDenied @nbouscal

Answer 4

unless you want to do this with calculus

Answer 5

its mechanics

Answer 6

@ParthKohli

Answer 7

@dpaInc @A.Avinash_Goutham @jhonyy9 @lgbasallote @Omniscience @myininaya

Answer 8

@UnkleRhaukus @Callisto @lgbasallote @nphuongsun93

Answer 9

\(s=804m\) , \(u=30ms^{-1}\) , \(a_1=-\frac{1}{2}ms^{-2} = -0.5ms^{-2} \) , \(a_2=-\frac{3}{2}ms^{-2}=-1.5ms^{-2} \)

Let \(v_1\) be the speed of car at the moment when deceleration is increased.
Consider the first part of the journey,
By \(v^2 = u^2 + 2as\)
\(v_1 \ ^2 = 30^2 + 2(-0.5)s_1\)
\(s_1 = 30^2 -v_1^2\)

At the end of the journey, the speed of car = 0ms^-1 (since it stopped)
Consider the second part of the journey,
By \(v^2 = u^2 + 2as\)
\(0 ^2 = v_1 \ ^2 + 2(-1.5)s_2\)
\(s_2 = \frac{v_1^2}{3}\)

Consider the whole journey,
\(s_1 + s_2 = 804\)
\((30^2 -v_1^2) + (\frac{v_1^2}{3}) = 804\)
\(\frac{-2}{3}v^2 = 804-900\)
\(v_1^2 = 144\)
\(v_1 = 12ms^{-1}\) or \(v_1 = -12ms^{-1}(rejected)\)

By v=u+at
Consider the first part of the journey,
\(v_1 = 30 + (-0.5)t_1\)
\(12 = 30 -0.5t_1\)
\(t_1=36s\)
Consider the second part of the journey,
\(0 = v_1 + (-1.5)t_2\)
\(0 =12-1.5t_2\)
\(t_2=8s\)
Total time = \(t_1 + t_2\) = 36s + 8s = 44s