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

(a) Total time = 105.5 sec

(b)average speed = 31.26 m/sec

Step-by-step 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* 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+(1)/(2)at^2=0* 17.5+(1)/(2)* 2* 17.5^2=306.25m`

Distance traveled in 83 sec with with velocity of 35 m/sec =
`35* 83=2905m`

For next 5 second

Initial velocity u = 35 m/sec

Final velocity = 0 as finally car stops

So
`a=(v-u)/(t)=(0-35)/(5)=-7m/sec^2`

Distance traveled
`s=ut+(1)/(2)at^2=35* 5+(1)/(2)* -7* 5^2=87.5m`

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

Total time = 105.5 sec

So average speed
`=(total\ distance)/(total\ time )=(3298.75)/(105.5)=31.26m/sec`