At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.50 m/s2. At the same instant, a truck, traveling with a constant speed of 14.8 m/s, overtakes and passes the automobile.

Part A
How far beyond its starting point does the automobile overtake the truck?

= m?

part B
How fast is the automobile traveling when it overtakes the truck?

= m/s ?

(A) Distance taken by automobile to overtake truck is 175.23 m

(B) Final speed of the automobile is 29.6 m/s.

Explanation:

According to the problem, the automobile is at rest, so its initial speed is zero and the automobile is moving with constant acceleration.

The general equation of motion is:

d = ut + 0.5at²      .....(1)

Here d is distance, a is acceleration, t is time and u is initial speed.

The distance (d₁) travel by the automobile after green light in time t is given by the relation :

d₁ = u₁t + 0.5at²

We know that its initial velocity is zero i.e. u = 0 m/s.

d₁ = 0.5at²      ......(2)

The truck is moving with constant speed, so its acceleration is zero. Therefore, the distance covered by the truck using equation (1) is;

d₂ = u₂t         .......(3)

Since, the distance travelled by automobile and truck are same. Hence,

d₁ = d₂

0.5at² = u₂t

t = 2u₂/a

Substitute 14.8 m/s for u₂ and 2.50 m/s² for a in the above equation.

t =

t = 11.84 s

(A) Distance travelled by automobile, d = u₂t =14.8 x 11.84 = 175.23 m

(B) Final speed of the automobile is given by the relation:

v = u + at

v = 0 + 2.50 x 11.84

v = 29.6 m/s

f=ma newton's second law 7x8=56n

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