[ALI ALZAHRANI]

المشاركة الأصلية كتبت بواسطة CDTOE أستاذ , علي .. لدي 4 مسائل جديدة تخص الحركة في خط أفقي , حتى الآن لم اقم بحل ايا منها لانشغالي بدراسة مواد أخرى , فاتمنى منك حلها : 1 - Spotting a police car , you brake a Porsche from 75 km/h to 45 km/h over a displacement of 88 m . a - what's the acceleration , assumed to be constant ? b - what's the elapsed time ? c - if you continue to slow down with the acceleration calculated in ( a ) , how much time will elapse in bringing the car to rest from 75 km/h ? d - In ( C ) , what's distance will be covered ? ××× 2 - An automobile travels with constant acceleration is brought to stop from a speed of 20 m / s in a distance of 40 m . a - What's acceleration ? b - what's the time it takes to stop the car ? ×××× 3 - A race car is moving at a constant speed of 35 m / s , passes stationary police car . The police car starts moving with a constant acceleration of 4 m/s^2 . Find the distance at which the police car overtakes the race car ? ××× 4 - A motorcyclist heading east through a small town accelerates as he passes the signpost at X=0 m marking the city limits . His acceleration is constant a = 4 m/sec^2 . At time t = 0 he's at X=5 m and his velocity V=3 m/sec a - find his position and velocity at time t=2 sec b - Where's he when his velocity is 5 m/sec ? ××××××× شكرا مقدما , وراح اخبرك بمطابقة حلي لحلك لاحقا عندما اتفرغ لها .

Since problems 1 and 2 (3 and 4) are very similar and can be solved using the same technique, I'll just pick one of each pair and you should solve the others

Q1: (a) It is known that

v^2 = v0^2 +2ax

Before going any further, you have to unify all the units in one similar kind of systems. Let's say we convert the km into m

v= 45km/hr = 12.5 m/s
v0= 75km/hr = 20.83 m/s

Therefore we get

a= (v^2 -v0^2)/2x = -1.58 m/s^2

(b) The ellapsed time can be estimated using the first equation of motion, that is

v = v0 +at

Now substitute all the known quatities into the previous equation to find the unknown, we find

t = (v -v0)/a = (12.5 -20.83)/-1.58 = 5.3 s

(c) To find the time taken for the car to come at rest from 75 km/hr is

t = (v -v0)/a

but note that the final speed is 0 and initial one is 20.83 m/s, therefore we find

t= (0 -20.83)/-1.58 = 13.18 s

(d) the distance travelled is

x = 0.5(v+v0)t = 0.5*(20.83+0)*13.18=137.3 m


Q3: Both cars will cover the same distance within the same piece of time, then

x(car) = x(police)

the car is moving with constant velocity that indicates that its acceleration is zero

x(car) = v(car) t=vt

the police car starts its motion from res which means its initial speed is zero

x(police) = 0.4 at^2
From the both equations we get

vt = 0.5 at^2 =============> v = 0.5 at

t = 2v/a = 2*35/4 =17.5 sec

Therefore the required distance is

x =vt = 35 * 17.5 =612.5 m = 0.612 km