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Chapter 1: Q34E (page 61)

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 2.80 m/s2. At the same instant a truck, travelling with a constant speed of 20.0 m/s, overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car travelling when it overtakes the truck? (c) Sketch an x-t graph of the motion of both vehicles. Take x = 0 at the intersection. (d) Sketch a vx-t graph of the motion of both vehicles.

Short Answer

Expert verified

(a) The total distance = 308.3 m.

(b) Speed of the car = 41.59 m/s.

Step by step solution

01

Identification of the given data

  • Constant acceleration (a) = 2.80 m/s2
  • Truck constant speed (v) = 20.8 m/s
02

calculation for the distance

(a)

The distance is calculated by using,

For the truck d=v脳t 鈥︹�︹�︹�︹�︹�︹��.. (1)

And also for the car d=v0+12脳a脳t2鈥︹�︹�︹��.. (2)

Comparing equations (1) and (2) we get,

v脳t=v0脳t+12脳a脳t2鈥︹�︹�︹�︹�︹�︹��.. (3)

Putting the given values in equation (3),

20.8脳t=0+12脳2.80脳t2t=2脳20.8m/s2.80m/s2t=14.85s

Then the distance will be,

d=v脳td=20.8m/s脳14.85s"d=308.8"m

Thus, the total distance will be d =308.8 m

03

calculation for the speed of the car

(b)

Velocity is calculated by using,

v=a脳t

Substituting the value

v=2.80m/s2脳14.85sv=41.59m/s

Thus, the speed of the car v=41.59 m/s .

04

x-t graph

(c)

05

Vx-t graph

(d)

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