+0  
 
0
569
2
avatar+4 

We have a car with a start velocity on 78 km/h. The car drives 345 meter and then the cars celocity is 45 km/h. The acceleration is constant.

What is the acceleration?
How long does it take the car to drive the distance?

 Oct 23, 2014

Best Answer 

 #2
avatar+27558 
+5

To check geno's results here's an alternative approach:

 

Acceleration:  use v2 = u2 + 2as where v is final velocity, u is initial velocity, a is acceleration, s is distance.

$${\mathtt{a}} = {\frac{\left({{\mathtt{45}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{78}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.345}}\right)}} \Rightarrow {\mathtt{a}} = -{\mathtt{5\,882.608\: \!695\: \!652\: \!173\: \!913}}$$  km/hr2

So 

a = -5882.608*1000/36002 m/s2 = -0.4539 m/s2

 

Time taken: use distance/average speed

average speed = (45 + 78)/2 km/h = 61.5 km/hr

time taken = 0.345km/61.5km/hr*3600s/hr = 20.195 seconds

 

Any differences from geno's results are just due to small numerical rounding errors, as his approach was perfectly ok.

.

 Oct 24, 2014
 #1
avatar+17746 
+5

I'm a rank amateur at physics; I'll try; but if others post, trust them!

78 km/hr = 21.7 m/s          45 km/hr = 12.5 m/s

Since the acceleration is constant:  acc = (final velocity - initial velocity) / time

    acc = (12.5 - 21.7)km/s / t sec  =  23.3 km / t sec²

distance = .5at² + (initial velocity)t

345 m = .5( 23.3 / t)t² + (21.7 m/sec)t

345 m = -4.6t  + 21.7t 

345 m = 17.1 t

t =  20.2 sec

a =  ( 12.5 m/s - 21.7 m/s ) / 20.2 s

a = -0.46 m/s²

 Oct 24, 2014
 #2
avatar+27558 
+5
Best Answer

To check geno's results here's an alternative approach:

 

Acceleration:  use v2 = u2 + 2as where v is final velocity, u is initial velocity, a is acceleration, s is distance.

$${\mathtt{a}} = {\frac{\left({{\mathtt{45}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{78}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.345}}\right)}} \Rightarrow {\mathtt{a}} = -{\mathtt{5\,882.608\: \!695\: \!652\: \!173\: \!913}}$$  km/hr2

So 

a = -5882.608*1000/36002 m/s2 = -0.4539 m/s2

 

Time taken: use distance/average speed

average speed = (45 + 78)/2 km/h = 61.5 km/hr

time taken = 0.345km/61.5km/hr*3600s/hr = 20.195 seconds

 

Any differences from geno's results are just due to small numerical rounding errors, as his approach was perfectly ok.

.

Alan Oct 24, 2014

41 Online Users

avatar
avatar
avatar
avatar