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Grade: 11

                        

A truck travelling along a straight road with a constant speed of 72 km/hr passes a 0 moving much slower. At the instant the truck passes the car, the car at time t car starts accelerating with constant acceleration 1m/s and overtakes the truck 0.6km further down the road The car moves with uniform velocity from this instant onwards. Find the distance between them at time t 50 sec

4 years ago

Answers : (2)

Shaswata Biswas
132 Points
							
Since in the question it is not mentioned about the initial velocity of the car, let it be at rest. The velocity of truck,  .   V= 20 m/s. 
Let the meet after t seconds.
Initial velocity of car, u= 0, acceleration, a = 1m/s
Distance travelled in t sec, S= at2/2
Distance travelled by truck, S2 = vTt
Then they are equal.
So, 20.t = 1.t2/2     => t=40 s
 
Velocity attained by the csr in 40 s, v= at = 1.40 = 40 m/s.
From hence, relative velocity if car w.r.t the truck, VCT = v- vT = 40 - 20 = 20 m/s.
So from the starting, distance between them is VCT.t = 20*30 = 600 m
4 years ago
Shaswata Biswas
132 Points
							
Or otherwise, let the initial velocity of the car be u. The velocity of truck,  .   V= 20 m/s. 
Let the car takes over the truck after t seconds
In this time,
Distance travelled by truck, S2 = vTt     => 600 = 20.t      => t = 30 s
So, the car takes over the truck after 30 seconds  
And, distance travelled by the car, S = u.t + at2/2     => 600 = 30.t + 302/2.    => u = 5 m/s.
 
Velocity attained by the csr in 30 s, v= u + at = 5 + 1.30 = 35 m/s.
From hence, relative velocity if car w.r.t the truck, VCT = v- vT = 35 - 20 = 15 m/s.
So, at t = 50 s, the car is ahead of the truck for (50-30) = 20 s
Hence, the distance between them is X = VCT.20 = 15*20 = 300 m
THANKS
4 years ago
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