A car slows down from 108 km/hr to 18 km/hr over a distance of 40 meter. If the brakes are applied with the same force . Calculate 

1) total time in which car comes to rest .

2) total distance travelled by it.

Asked by vishwadiprathod0070.9sdatl | 7th May, 2020, 11:10: AM

Expert Answer:

When, u = 108 km/h = 30 m/s
and v = 18 km/h = 5 m/s
s = 40 m 
 
We can calculate the acceleration using 3rd equation of motion: - 
 
v2 - u2 = 2as 
(5)2 - (30)2 = 2 a (40) 
 
25 - 900 = 80a
a = - 10.93 m/s2 
Negative sign indicates that it is retardation. 
 
i)Total time in which car comes to rest: - 
As the brakes are applied with same force to bring car to rest, acceleration remains same i.e. 10.93 m/s2. 
 
v = 0, u = 30 m/s 
a = -10.93 m/s2 
 
Thus, 
v = u + at 
0 = 30 + (-10.93) t 
-30 = -10.93 t 
Thus, 
t =  2.74 s
 
ii) Distance travelled by car during this time is given by, 
 
v2 - u2 = 2as 
(0)2 - (30)2 = 2(-10.93) s
 
- 900 = 21.86 s 
s =  41.17 m 
 
 
 
 

Answered by Shiwani Sawant | 7th May, 2020, 05:52: PM