21-Q21- An electric car starts from rest and accelerates at the rate of 2 m-s2 nbsp- in straight line until it reaches a speed of 20 m-s- Now the car slows at constant rate 1 m-s2 until it stops- Total distance from starts to stop is- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -A- 300 m nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-B- 200 m- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -C- 100 m nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-D- 500 m

Dear Student , Vf = at + Vi nbsp; Vf = final velocity (20 m/s) nbsp; a = acceleration (2 m/s2) nbsp; t = time nbsp; Vi = initial velocity (0 m/s) nbs ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Finding Average Velocity

21 Q21. A...

Question

21

Q21. An electric car starts from rest and accelerates at the rate of 2 m/s² in straight line until it reaches a speed of 20 m/s. Now the car slows at constant rate 1 m/s²until it stops. Total distance from starts to stop is

(A) 300 m (B) 200 m

(C) 100 m (D) 500 m

Open in App

Solution

Dear Student ,
Vf = at + Vi

Vf = final velocity (20 m/s)
a = acceleration (2 m/s²)
t = time
Vi = initial velocity (0 m/s)

20 = 2t + 0
t = 10s

Find the distance as the car accelerates
Xf = 0.5at² + Vt + Xi

Xf = final position
Xi = initial position
Xf = 0.5(2)(10)²
Xf = 110m

Vf = at + Vi

now that the car decelerates, its initial velocity at the time is 20 m/s. When the car stops, its final velocity is 0 m/s
0 = -1t + 20
-20 = -t
t = 20 s

find the distance as it decelerates
Xf = 0.5at² + Vt + Xi
Xf = 0.5(-1)(20)² + 20(20)
Xf = 200 m

the total time is 10 + 20 = 30 s
the total distance is 200 + 110 = 310 m

Regards

Suggest Corrections

Similar questions

Q. A car starting from rest, is accelerated at a constant rate

α

until it attains a speed

V

. It is then retarded at a constant rate

β

until it comes to rest. The average speed of the car during its entire journey is :

Q. A car starts from rest and acceleration at

4 {m / s}^{2}

for

5 s

. After that it moves with constant velocity for

25 s

and then retards at

2 m / s^{2}

until it comes to rest. The total distance travelled by the car is

Q. A car starting from rest accelerates at the rate

f

through a distance

s_{1}

, and then decelerates at rate

\frac{f}{2}

through a distance

s_{2}

to come to rest. Then

\frac{s_{1}}{s_{2}}

Q. A car starting from rest is accelerated at constant rate until it attains a constant speed,

v

. It is then retarded at a constant rate until it comes to rest. Considering that the car moves with constant speed for half of the time of total journey, the average speed of the car for the journey is

A car, starting from rest, is accelerated at a constant rate $α$ until it attains a speed $v$ . It is then retarded at a constant rate $β$ until it comes to rest. The average speed of the car during its entire journey is

Join BYJU'S Learning Program

21 ​Q21. An electric car starts from rest and accelerates at the rate of 2 m/s2 in straight line until it reaches a speed of 20 m/s. Now the car slows at constant rate 1 m/s2 until it stops. Total distance from starts to stop is (A) 300 m (B) 200 m (C) 100 m (D) 500 m

21

Q21. An electric car starts from rest and accelerates at the rate of 2 m/s² in straight line until it reaches a speed of 20 m/s. Now the car slows at constant rate 1 m/s²until it stops. Total distance from starts to stop is

(A) 300 m (B) 200 m

(C) 100 m (D) 500 m