AS Physics past paper questions?

Question

A vechicle on a straight downhill road accelerated uniformly from a speed of 4.0 ms-1 to a speed of 29ms-1 over a distance of 850 m, when the driver braked and stopped the vechicle in 28s.

a) Calculate the time taken to reach 29 ms-1 from 4 ms-1.

ii) Calculate its acceleration during this time.

b) Calculate the distance it travelled during deceleration.

ii) Its average deceleration as it slowed.

AS Physics past paper questions, I need help, thanks.

Answer 1

V=V0 + at where
V= speed of the vehicle
V0 - initial speed of the vehicle
a - acceleration
t - time

Also
S=V0t +0.5 at^2Where
S- distace covered

from the first equation
a=(V-Vo)/t

then
S=V0t + 0.5 [(V-Vo)/t]t^2
S=[V0 + 0.5 (V-Vo)] t

a) t=S/(V0 + 0.5 (V-Vo)) or
t=850 /(4 + 0.5 (29 - 4))
t=51.5sec

ii) a = (V-V0)/t
a= (29-4)/51.5=.49 m/s^2

b) S=V0 -.5at^2
a=V0/t

ii)a=V0/t

Answer 2

Where are you having trouble? This is a straightforward time & motion problem. The acceleration is constant, you can compute it as (65m/s) / 25s. Given that, you can use the distance equation d = 0.5at^2 + vt + c to compute the distance traveled. Since the initial position and velocity are both 0, this is simply 1/2 a t^2. Does that do it?