Question

7. A car starts from rest and accelerates at 3 m/s2 for 5 seconds and then continues with constant velocity. Calculate the distance covered in 15 seconds since it starts from rest.

Answer 1

The distance travelled by the car in first 5 seconds

S=ut+1/2at²

u=0as the car starts from rest

S=1/2×3×25

S=37.5

The final velocity of the first part will be the initial velocity of the second part

V=at

V=15 m/s

The distance covered in the next 10 seconds

S1=V×t=10×15=150

Total distance covered

=150+37.5=187.5

I hope this helps ( ╹▽╹ )

Answer 2

Answer:

$37.5 \: m$

Explanation:

$Initial \: vel \: (u) \: = 0$

$Acceleration \: (a) \: = 5 {m/s}^{2}$

$Time \: (t) \: = 5s$

$Therefore, \\ Final \: vel \: (v) \: = u + at \\ = 0 + 3 \times 5 \\ = 15 \: m/s$

$Therefore, \\ Distance(s) \: = \: u + \frac{1}{2} {at}^{2} \\ = 0 + \frac{1}{2} \times 3 \times {5}^{2} \\ = \frac{1}{2} \times 3 \times 25 \\ = \frac{1}{2} \times 75 \\ = 37.5 \: m$

$Hope \: this \: will \: help \: u....$