Question

1
4 The diagram sh
ows a velocity-time graph for a car from the time the driver sees an
obstacle in the road until the car comes to rest.
30
Velocity
in m/s
20-
1
10-
0-
0
2.0
4.0
6.0 8.0
Time in seconds
(a) (i) Calculate the acceleration of the car between 1.8 and 8.0 seconds.
- 1xbxh
₃ x 20 x 30
= 300
acceleration =
(ii) Calculate the braking distance of the car.
Thonking
+ breakin​

Answer 1

The braking distance is 92.9752m/s and the acceleration is -4.84m/s^2.

Given in the graph that,

Initial velocity(u)=30m/s

Final velocity(v)=0m/s

Initial time($t_{2}$)=1.8s

Final time($t_{1}$)=8.0s

i) To find acceleration,

accelaration= $\frac{v-u}{t_{2} -t_{1} }$

                    =$\frac{0-30}{8.0-1.8}$

                    =$\frac{-30}{6.2} =-4.84m/s^{2}$

The negative sign in the acceleration shows that the vehicle is retarding.

ii)To find the braking distance (S)

    Time taken to change velocity from 30m/s to 0m/s,

       t=8.0*1.8

        =6.2s

  Braking distance (S)= $ut+\frac{1}{2}at^{2}$

                                   $=30*6.2-\frac{1}{2}*4.84*6.2^{2} \\=92.9752m$

Therefore the acceleration between 1.8s and 8.0s is 4.84m/s^2

and the braking distance is 92.9752m

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