Question

a car starting from rest attains a velocity of 60km/h in 1min.calculate :1)its average velocity 2)acceleration 3)distance travelled in first 1min

Answer 1

2. v=60km/h
60*5/18 =16.66m/s
u=0
t = 1*60=60s
v=u +at
16.66= 0+a*60
16.66/60=a
0.277m/s square
3 s= ut+at^2
s= 0*60+0.277*60^2
0+ 997.2
s= 997.2

Answer 2

Answer:

The average velocity is 8.33 m/s, the acceleration is 0.278 m/s² and the distance traveled in first 1 min is 999.6 m.

Explanation:

Given that,

Velocity v = 60 =16.66 m/s

Time t = 1 min = 60 sec

(I). The average velocity is

$v_{av}=\dfrac{v_{i}+v_{f}}{2}$

$v_{av}=\dfrac{0+16.66}{2}$

$v_{av}=8.33\ m/s$

(2). The acceleration is

Using equation of motion

$v = u+at$

$a =\dfrac{v-u}{t}$

$a=\dfrac{16.66}{60}$

$a = 0.278\ m/s^2$

(3). The distance traveled in first 1 mint

Using the distance formula

$d = v\times t$

$d = 16.66\times60$

$d =999.6\ m$

Hence, The average velocity is 8.33 m/s, the acceleration is 0.278 m/s² and the distance traveled in first 1 min is 999.6 m.