Question

A three-wheeled car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 35.0 m/s. Then the vehicle moves for 83.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the three-wheeled car in motion (in s)? 105.5 s (b) What is the average velocity of the three-wheeled car for the motion described? (Enter the magnitude in m/s.) m/s

Answer 1

Answer:

(a) Total time = 105.5 sec

(b)average speed = 31.26 m/sec

Explanation:

We have given that three-wheeled car starts from rest that initial velocity u = 0 m/sec

Final velocity = 35 m/sec

Acceleration $a=2m/sec^2$

According to first law of motion we know that v = u+at

So $35=0+2\times t$

t = 17.5 sec

After this car travels 83 sec at a constant speed and after that it takes 5 sec additional time to stop

(a) So total time in which car is in motion = 17.5+83+5 = 105.5 sec

(b) Total distance traveled during first 17.5 sec

$s=ut+\frac{1}{2}at^2=0\times 17.5+\frac{1}{2}\times 2\times 17.5^2=306.25m$

Distance traveled in 83 sec with with velocity of 35 m/sec = $35\times 83=2905m$

For next 5 second

Initial velocity u = 35 m/sec

Final velocity = 0 as finally car stops

So $a=\frac{v-u}{t}=\frac{0-35}{5}=-7m/sec^2$

Distance traveled $s=ut+\frac{1}{2}at^2=35\times 5+\frac{1}{2}\times -7\times 5^2=87.5m$

So total distance traveled = 306.25+2905+87.5 = 3298.75 m

Total time = 105.5 sec

So average speed $=\frac{total\ distance}{total\ time }=\frac{3298.75}{105.5}=31.26m/sec$