St.
Theresa School (Dire Dawa)
Physics Tutorial for Grade-12 students
Equations of Uniformly Accelerated Motion in 1D
v = u + at (1)
u+v
s=( )t (2)
2
1
s = ut + at2 (3)
2
v 2 = u2 + 2as (4)
1
s = vt − at2 (5)
2
Determining the stopping distance of a vehicle
When the body is moving with a certain velocity and suddenly brakes are applied, you would
have noticed that the body stops completely after covering a certain distance. This is called
the stopping distance.
• The stopping distance is the distance travelled between the time when the body decides
to stop a moving vehicle and the time when the vehicle stops completely.
• The stopping distance depends on factors including road surface, and reflexes of the car’s
driver and it is denoted by d.
There are many factors that can affect the stopping distance of a moving vehicle.
• Speed of the vehicle
• Weight of the vehicle
• Road conditions (slick, icy, snow, dry, wet)
• Vehicle brake conditions (old or worn pads and rotors)
• Braking technology in the vehicle
• Tire conditions
Stopping distance = reaction distance + braking distance
Reaction distance is the product of the car’s speed and the reaction time, and is given by
Sr = vt (6)
• The reaction distance is the distance you travel from the point of detecting a hazard
until you begin braking.
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Prepared By: Abdi B. (M.sc) November 2024
St. Theresa School (Dire Dawa)
Physics Tutorial for Grade-12 students
• It is the distance the vehicle travels while the driver reacts to the hazard and applies the
brakes.
• Braking distance is the distance the vehicle travels after the driver applies the brakes.
Braking distance formula is given by
v2
Sb = (7)
2µg
, where d is stopping distance (m), v is velocity (m/s), µ friction coefficient and g is acceleration
due to gravity.
The stopping distance formula is also given by
Sb = kv 2
1
, where k = 2µg is the proportionality constant. Using equations (6) and (7), the stopping
distance of a vehicle becomes
v2
d = Sr + Sb = vt + (8)
2µg
Examples
1. A race car was is moving at a constant speed of 35 m/s. A security car was moving at
a speed of 5 m/s as the race car passes by it and was accelerating at constant rate of 5
m/s2 . What was the speed of the security car when it took over the race car?
2. At a speed of 20 m/s, a particular vehicle had a stopping distance of 40 metres. The car
travelled 14metres whilst the driver was reacting to the incident in front of him. What
was the braking distance?
3. A bike moves with a velocity of 15 m/s and applies a brake. Calculate its braking distance
if the constant of proportionality is 0.9.
4. A vehicle is travelling 105 km/h on a highway when the driver sees a moose in the middle
of the lane.
(a) Determine the distance travelled by the vehicle while the driver is reacting, assume
the driver’s reaction time is 1.50 s.
(b) Determine the braking distance, assuming a deceleration of 5.85 m/s2.
(c) Determine the stopping distance.
5. A car is moving with a velocity of 40 m/s and suddenly applies brakes. Determine the
constant of proportionality if the body covers a distance of 10 m before coming to rest.
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Prepared By: Abdi B. (M.sc) November 2024
St. Theresa School (Dire Dawa)
Physics Tutorial for Grade-12 students
6. While driving at a speed of 35 m/s, Stephen sees an obstacle in the road at time t = 0
.The velocity-timegraph below shows how the speed of the car changes as Stephen reacts
and slams the brakes,bringing the car to a halt. Determine
(a) The braking distance of the car
(b) The driver’s reaction time
(c) The coefficient of friction between the road and the tire of the car.
7. The braking distance of a typical car of mass 1500 kg is about 24 m and its kinetic energy
is 168 750 J.
(a) Estimate the coefficient of friction between the road and the tyre of the car
(b) If the reaction time of the driver is 1.50s, what is the stopping distance of the car?
8. A car has an initial velocity v0 when the driver sees an obstacle in the road in front of
him. His reaction time is tr , and the braking acceleration of the car is a. Show that the
total stopping distance is given by
v02
Sstop = v0 tr −
2a
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Prepared By: Abdi B. (M.sc) November 2024
