Question

A car starts from rest and accelerates at a constant rate travelling a distance of 210 m in 14.9 s. The car continues to accelerate at the same rate for an additional 199 m. What is the speed (in m/s) of the car after it has travelled the (210+199) m?

Answer 1

Under uniform acceleration,s=ut+1/2a*t^2,where s is displacement, u is initial velocity, t is time and a is acceleration.

Here,s=210 m, t=14.9 s,u=0 m/s(the car starts from rest)

So,210=0*14.9+1/2*a*14.9*14.9

=> a=210*2/(14.9*14.9)=1.8918 m/s2.

Also, under uniform acceleration, v^2=u^2+2*a*s, where v is final velocity, u is initial velocity, a is acceleration and s is displacement.

For the complete motion,s=210+199 m=409 m, u= 0 m/s, a=1.8918 m/s2.

So,v^2=0^2+2*1.8918*409=1547.4924

=> v=39.34 m/s

So, required velocity=39.34 m/s