# Car at 79km/hr hits brakes and comes to stop in 11.3 seconds. What was car’s deceleration and how far did it go before it came to a stop?

### 4 Answers

- NCSLv 78 months agoFavourite answer
79km/h = 21.94 m/s

acceleration a = -21.94m/s / 11.3s = -1.94 m/s²

and for "deceleration" you probably don't need the sign.

distance s = Vavg * t = ½ * 21.94m/s * 11.3s = 124 m

or

s = v0+t + ½at²

or

v² = 0 = u² + 2as

all give the same result

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- MyRankLv 67 months ago
Given,

Time (t) = 11.3sec

Deceleration (a) = ?

Distance (d) = ?

Final speed (u) = 0m/sec

Initial speed (v) = 79km/hr = 21.94m/sec

Using kinematic relation:-

Acceleration (a) = v-u/t

= (0-21.94)m/sec / 11.3sec = -21.94/11.3 m/sec²

= -1.94m/sec²

∴ Acceleration (a) = -1.94m/sec²

Now, v²= u²+2ad

(0m/sec)² = (21.94m/sec)² + 2(-1.94m/sec²) x d

= (21.94)²m²/sec² = 3.88m/sec² x d

= 481.3636m²/sec² = 3.88m/sec² x d

= Distance (d) = 481.3636/3.88 = 124m

∴ Distance (d) = 124m.

Source(s): https://myrank.co.in/ - Anonymous8 months ago
Unanswerable regarding travel distance.

Even Courts & Police have a difficulty in establishing "skid" distance re speed.

Tyre Friction, and road friction coefficient is nearly a science unto itself.

Damp pavement, Icy pavement, concrete, MacCadam, gravel, tar, leaves, sand, etc.

- Anonymous8 months ago
acceleration = change in speed / time = 6.99 km / hr-sec

distance = 1/2 * acceleration * time^2