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just as a traffic light turns green, a waiting car starts off with a constant acceleration of 6.0 m/s2 . at the instant the car begins to accelerate, a truck with a constant velocity of 21 m/s passes in the next lane. (a) how far will the car travel before it overtakes the truck? (b) how fast will the car be traveling when it overtakes the truck?



Answer :

(a) The car travels a distance of 147 m before it overtakes the truck.

(b) At the moment when car overtakes the truck, the speed of car was 21 m/s

Let us assume that the time taken by the car to overtake the truck = t seconds

Now calculating the distance travelled by car and truck during this time,

Using the kinematic equation for calculating the distance travelled by car and the truck:

The distance travelled can be computed by the relationship [tex]s= 0.5at^2[/tex]

Distance travelled by car will be [tex]S_{car} =0.5 \times 6\times t^2=3t^2 \:m[/tex]

Distance travelled by truck will be [tex]S_{truck}=21 \times t = 21t \:m[/tex]

At the instant of overtaking

Distance travelled by car = Distance travelled by truck

[tex]3t^2 = 21t\Rightarrow t = \frac{21t}{3t}=7\:sec.[/tex]

So, the distance travelled by car before it overtakes will be 3*7*7 = 147 m

(b) As Car have crossed the truck in 7 sec. after covering a distance of 147 m.

So, at the moment of overtaking the truck the speed of the car will be = 147/7 = 21/s

For more questions on kinematic equation

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