Question

04. A car accelerates from rest at a constant rate 10 ms 2 for some time after which it decelerates at a constant rate
8 ms2 to come to rest. If the total time lapse is 60 s, find the
(i) Maximum velocity attained
(ii) The total distance travelled.

Answer 1

Explanation:

let time in which it accelerates be (t)s

s1 be distance covered in (t)s

so time in which it decelarates is (60-t)s

s2 be distance travelled in (60-t)s

[case - 1]

u = 0 m/s

a = 10 m/s²

v = u + at => 0 + 10t => 10t m/s

s1 = ut + 1/2at² = 5t²

[case - 2]

u = 10t m/s

a = -8 m/s²

v = 0 m/s

v = u + at => 0 = 10t + (-8)(60-t) => t = 80/3 s

s2 = ut + 1/2at² = 10t(60-t) + (-4)(60-t)²

= 600t - 10t² - 240 + 4t²

= -6t² + 600t - 240

= 40880/3 metres

s1 = 32000/3 metres

total distance = s1 + s2 = 72880/3 metres = 24293.33 metres

max velocity = 800/3 m/s = 266.66 m/s