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Grade: 11

                        

A car accelerates from rest at a constant rate a for some time ,after which it decelerates at a constant rate b to come to rest .If total time elapsed is t second ,then calculate : (i) the maximum velocity attained by the car (ii)the total distance travelled by the car in terms of a,b and t.

3 years ago

Answers : (3)

Arun
24737 Points
							
Let the car accelerate for t₁sec at a m/sec² and decelerate for t₂sec. at - b/ sec² 
At the end of t₁ sec. its velocity = 0 + at₁ m/sec. At the end of t₂ sec its velocity = 0 
Initial velocity at the beginning of deceleration is α t₁ 
V(t₂) = a t₁ - b t₂ = 0 Therefore t₂ = (a/ b ) t₁ ------------------------(1) 
But t = t₁ + t₂ ⇒ t = t₁ + t₂ = t₁ + (a/ b ) t₁ = t₁(1+ a/ b) 
∴ t₁ = t / (1+ a/ b) = {b / (a + b)}*t -------------------------------------- (2) 
Max. Velocity reached = a t₁ = a t / (1+ a/ b) = t / (1/ a + 1 / b) 
 = a*b*t/(a+b)
Distance covered =(a t₁² + b t₂²) /2 = ={a t₁² + b [(a/ b ) t₁)²]} /2 (substitute for t₂ from (1) 
Simplifying this = ½ * (a/ b) * (a + b)* t₁²
Now substitute for t₁ from (2) to get 
Distance covered = ½ *[a*b / (a +b)]*t²
3 years ago
Vimansh
13 Points
							
Total time equals T 
a=v/t1
t1=v/a......1
t1 + t2 = T
t2=T-t1
B=u/T-t1
B=v/T-t1 (v in first case equals u in second case).....2
T=v/B+v/a
T=v(1/B + 1/a)
T=v(a+B/aB)
v=(aB/a+B)T
a is alpha and B is for beta
one year ago
ankit singh
498 Points
							
For the first leg of the journey,
V = 0 + α  t1
So  V = α  t1
for the second leg of journey of duration  t - t1 :
0 = V - β (t - t1)
0 = α t1 - β t + β t1
t1 = β t / (α + β)
V =  αt1 = α β t / (α + β )
 
2 months ago
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